If A is an acute angle and sin A = 3 5 \dfrac{3}{5} 5 3
Answer
Given sin A = 3 5 \dfrac{3}{5} 5 3
sin2 A + cos2 A = 1
Putting values we get,
⇒ ( 3 5 ) 2 + cos 2 A = 1 ⇒ 9 25 + cos 2 A = 1 ⇒ cos 2 A = 1 − 9 25 ⇒ cos 2 A = 25 − 9 25 ⇒ cos 2 A = 16 25 ⇒ cos A = 16 25 ⇒ cos A = 4 5 . \Rightarrow \Big(\dfrac{3}{5}\Big)^2 + \text{cos}^2A = 1 \\[1em] \Rightarrow \dfrac{9}{25} + \text{cos}^2A = 1 \\[1em] \Rightarrow \text{cos}^2A = 1 - \dfrac{9}{25} \\[1em] \Rightarrow \text{cos}^2A = \dfrac{25 - 9}{25} \\[1em] \Rightarrow \text{cos}^2 A = \dfrac{16}{25} \\[1em] \Rightarrow \text{cos} A = \sqrt{\dfrac{16}{25}} \\[1em] \Rightarrow \text{cos }A = \dfrac{4}{5}. ⇒ ( 5 3 ) 2 + cos 2 A = 1 ⇒ 25 9 + cos 2 A = 1 ⇒ cos 2 A = 1 − 25 9 ⇒ cos 2 A = 25 25 − 9 ⇒ cos 2 A = 25 16 ⇒ cos A = 25 16 ⇒ cos A = 5 4 .
sec A = 1 cos A = 1 4 5 = 5 4 \dfrac{1}{\text{cos A}} = \dfrac{1}{\dfrac{4}{5}} = \dfrac{5}{4} cos A 1 = 5 4 1 = 4 5
cosec A = 1 sin A = 1 3 5 = 5 3 . \dfrac{1}{\text{sin A}} = \dfrac{1}{\dfrac{3}{5}} = \dfrac{5}{3}. sin A 1 = 5 3 1 = 3 5 .
1 + tan2 A = sec2 A
Putting values we get,
1 + tan 2 A = ( 5 4 ) 2 1 + tan 2 A = 25 16 tan 2 A = 25 16 − 1 tan 2 A = 25 − 16 16 tan 2 A = 9 16 tan A = 9 16 tan A = 3 4 . 1 + \text{tan}^2A = \Big(\dfrac{5}{4}\Big)^2 \\[1em] 1 + \text{tan}^2A = \dfrac{25}{16} \\[1em] \text{tan}^2A = \dfrac{25}{16} - 1 \\[1em] \text{tan}^2A = \dfrac{25 - 16}{16} \\[1em] \text{tan}^2A = \dfrac{9}{16} \\[1em] \text{tan }A = \sqrt{\dfrac{9}{16}} \\[1em] \text{tan }A = \dfrac{3}{4}. 1 + tan 2 A = ( 4 5 ) 2 1 + tan 2 A = 16 25 tan 2 A = 16 25 − 1 tan 2 A = 16 25 − 16 tan 2 A = 16 9 tan A = 16 9 tan A = 4 3 .
1 + cot2 A = cosec2 A
Putting values we get,
1 + cot 2 A = ( 5 3 ) 2 1 + cot 2 A = 25 9 cot 2 A = 25 9 − 1 cot 2 A = 25 − 9 9 cot 2 A = 16 9 cot A = 16 9 cot A = 4 3 . 1 + \text{cot}^2A = \Big(\dfrac{5}{3}\Big)^2 \\[1em] 1 + \text{cot}^2A = \dfrac{25}{9} \\[1em] \text{cot}^2A = \dfrac{25}{9} - 1 \\[1em] \text{cot}^2A = \dfrac{25 - 9}{9} \\[1em] \text{cot}^2A = \dfrac{16}{9} \\[1em] \text{cot }A = \sqrt{\dfrac{16}{9}} \\[1em] \text{cot }A = \dfrac{4}{3}. 1 + cot 2 A = ( 3 5 ) 2 1 + cot 2 A = 9 25 cot 2 A = 9 25 − 1 cot 2 A = 9 25 − 9 cot 2 A = 9 16 cot A = 9 16 cot A = 3 4 .
Hence, the value of,
cos A = 4 5 tan A = 3 4 cot A = 4 3 sec A = 5 4 cosec A = 5 3 . \text{cos A} = \dfrac{4}{5} \\[1em] \text{tan A} = \dfrac{3}{4} \\[1em] \text{cot A} = \dfrac{4}{3} \\[1em] \text{sec A} = \dfrac{5}{4} \\[1em] \text{cosec A} = \dfrac{5}{3}. cos A = 5 4 tan A = 4 3 cot A = 3 4 sec A = 4 5 cosec A = 3 5 .
If A is an acute angle and sec A = 17 8 \dfrac{17}{8} 8 17
Answer
Given sec A = 17 8 \dfrac{17}{8} 8 17
⇒ cos A = 1 sec A = 1 17 8 = 8 17 \dfrac{1}{\text{sec A}} = \dfrac{1}{\dfrac{17}{8}} = \dfrac{8}{17} sec A 1 = 8 17 1 = 17 8
sin2 A + cos2 A = 1
Putting values we get,
⇒ sin 2 A + ( 8 17 ) 2 = 1 ⇒ sin 2 A + 64 289 = 1 ⇒ sin 2 A = 1 − 64 289 ⇒ sin 2 A = 289 − 64 289 ⇒ sin 2 A = 225 289 ⇒ sin A = 225 289 ⇒ sin A = 15 17 . \Rightarrow \text{sin}^2A + \Big(\dfrac{8}{17}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2A + \dfrac{64}{289} = 1 \\[1em] \Rightarrow \text{sin}^2A = 1 - \dfrac{64}{289} \\[1em] \Rightarrow \text{sin}^2A = \dfrac{289 - 64}{289} \\[1em] \Rightarrow \text{sin}^2 A = \dfrac{225}{289} \\[1em] \Rightarrow \text{sin } A = \sqrt{\dfrac{225}{289}} \\[1em] \Rightarrow \text{sin }A = \dfrac{15}{17}. ⇒ sin 2 A + ( 17 8 ) 2 = 1 ⇒ sin 2 A + 289 64 = 1 ⇒ sin 2 A = 1 − 289 64 ⇒ sin 2 A = 289 289 − 64 ⇒ sin 2 A = 289 225 ⇒ sin A = 289 225 ⇒ sin A = 17 15 .
cosec A = 1 sin A = 1 15 17 = 17 15 . \dfrac{1}{\text{sin A}} = \dfrac{1}{\dfrac{15}{17}} = \dfrac{17}{15}. sin A 1 = 17 15 1 = 15 17 .
1 + tan2 A = sec2 A
Putting values we get,
1 + tan 2 A = ( 17 8 ) 2 1 + tan 2 A = 289 64 tan 2 A = 289 64 − 1 tan 2 A = 289 − 64 64 tan 2 A = 225 64 tan A = 225 64 tan A = 15 8 . 1 + \text{tan}^2A = \Big(\dfrac{17}{8}\Big)^2 \\[1em] 1 + \text{tan}^2A = \dfrac{289}{64} \\[1em] \text{tan}^2A = \dfrac{289}{64} - 1 \\[1em] \text{tan}^2A = \dfrac{289 - 64}{64} \\[1em] \text{tan}^2A = \dfrac{225}{64} \\[1em] \text{tan }A = \sqrt{\dfrac{225}{64}} \\[1em] \text{tan }A = \dfrac{15}{8}. 1 + tan 2 A = ( 8 17 ) 2 1 + tan 2 A = 64 289 tan 2 A = 64 289 − 1 tan 2 A = 64 289 − 64 tan 2 A = 64 225 tan A = 64 225 tan A = 8 15 .
1 + cot2 A = cosec2 A
Putting values we get,
1 + cot 2 A = ( 17 15 ) 2 1 + cot 2 A = 289 225 cot 2 A = 289 225 − 1 cot 2 A = 289 − 225 225 cot 2 A = 64 225 cot A = 64 225 cot A = 8 15 . 1 + \text{cot}^2A = \Big(\dfrac{17}{15}\Big)^2 \\[1em] 1 + \text{cot}^2A = \dfrac{289}{225} \\[1em] \text{cot}^2A = \dfrac{289}{225} - 1 \\[1em] \text{cot}^2A = \dfrac{289 - 225}{225} \\[1em] \text{cot}^2A = \dfrac{64}{225} \\[1em] \text{cot }A = \sqrt{\dfrac{64}{225}} \\[1em] \text{cot } A = \dfrac{8}{15}. 1 + cot 2 A = ( 15 17 ) 2 1 + cot 2 A = 225 289 cot 2 A = 225 289 − 1 cot 2 A = 225 289 − 225 cot 2 A = 225 64 cot A = 225 64 cot A = 15 8 .
Hence, the value of,
sin A = 15 17 cos A = 8 17 tan A = 15 8 cot A = 8 15 cosec A = 17 15 . \text{sin A} = \dfrac{15}{17} \\[1em] \text{cos A} = \dfrac{8}{17} \\[1em] \text{tan A} = \dfrac{15}{8} \\[1em] \text{cot A} = \dfrac{8}{15} \\[1em] \text{cosec A} = \dfrac{17}{15}. sin A = 17 15 cos A = 17 8 tan A = 8 15 cot A = 15 8 cosec A = 15 17 .
If 12 cosec θ = 13, find the value of 2 sin θ − 3 cos θ 4 sin θ − 9 cos θ \dfrac{2\text{ sin θ} - 3\text{ cos θ}}{4\text{ sin θ} - 9\text{ cos θ}} 4 sin θ − 9 cos θ 2 sin θ − 3 cos θ
Answer
Given 12 cosec θ = 13
⇒ cosec θ = 13 12 \dfrac{13}{12} 12 13
1 + cot2 θ = cosec2 θ
Putting values we get,
1 + cot 2 θ = ( 13 12 ) 2 1 + cot 2 θ = 169 144 cot 2 θ = 169 144 − 1 cot 2 θ = 169 − 144 144 cot 2 θ = 25 144 cot θ = 25 144 cot θ = 5 12 . 1 + \text{cot}^2\spaceθ = \Big(\dfrac{13}{12}\Big)^2 \\[1em] 1 + \text{cot}^2\spaceθ = \dfrac{169}{144} \\[1em] \text{cot}^2\spaceθ = \dfrac{169}{144} - 1 \\[1em] \text{cot}^2\spaceθ = \dfrac{169 - 144}{144} \\[1em] \text{cot}^2\spaceθ = \dfrac{25}{144} \\[1em] \text{cot }θ = \sqrt{\dfrac{25}{144}} \\[1em] \text{cot } θ = \dfrac{5}{12}. 1 + cot 2 θ = ( 12 13 ) 2 1 + cot 2 θ = 144 169 cot 2 θ = 144 169 − 1 cot 2 θ = 144 169 − 144 cot 2 θ = 144 25 cot θ = 144 25 cot θ = 12 5 .
We need to find the value of 2 sin θ − 3 cos θ 4 sin θ − 9 cos θ \dfrac{2\text{ sin θ} - 3\text{ cos θ}}{4\text{ sin θ} - 9\text{ cos θ}} 4 sin θ − 9 cos θ 2 sin θ − 3 cos θ
Dividing numerator and denominator of above expression by sin θ.
⇒ 2 sin θ − 3 cos θ sin θ 4 sin θ − 9 cos θ sin θ = 2 − 3 cot θ 4 − 9 cot θ = 2 − 3 × 5 12 4 − 9 × 5 12 = 2 − 5 4 4 − 15 4 = 8 − 5 4 16 − 15 4 = 3 4 1 4 = 3 × 4 4 = 3. \Rightarrow \dfrac{\dfrac{2\text{ sin θ} - 3\text{ cos θ}}{\text{ sin θ}}}{\dfrac{4\text{ sin θ} - 9\text{ cos θ}}{\text{sin θ}}} \\[1em] = \dfrac{2 - 3\text{ cot θ}}{4 - 9\text{ cot θ}} \\[1em] = \dfrac{2 - 3 \times \dfrac{5}{12}}{4 - 9 \times \dfrac{5}{12}} \\[1em] = \dfrac{2 - \dfrac{5}{4}}{4 - \dfrac{15}{4}} \\[1em] = \dfrac{\dfrac{8 - 5}{4}}{\dfrac{16 - 15}{4}} \\[1em] = \dfrac{\dfrac{3}{4}}{\dfrac{1}{4}} \\[1em] = \dfrac{3 \times 4}{4} \\[1em] = 3. ⇒ sin θ 4 sin θ − 9 cos θ sin θ 2 sin θ − 3 cos θ = 4 − 9 cot θ 2 − 3 cot θ = 4 − 9 × 12 5 2 − 3 × 12 5 = 4 − 4 15 2 − 4 5 = 4 16 − 15 4 8 − 5 = 4 1 4 3 = 4 3 × 4 = 3.
Hence, the value of the expression is 3.
Without using trigonometric tables, evaluate the following:
cos 2 26 ° + cos 64° sin 26° + tan 36° cot 54° . \text{cos}^2 \space 26° + \text{cos 64° sin 26°} + \dfrac{\text{tan 36°}}{\text{cot 54°}}. cos 2 26° + cos 64° sin 26° + cot 54° tan 36° .
Answer
We need to find the value of
cos 2 26 ° + cos 64° sin 26° + tan 36° cot 54° \text{cos}^2\space26° + \text{cos 64° sin 26°} + \dfrac{\text{tan 36°}}{\text{cot 54°}} cos 2 26° + cos 64° sin 26° + cot 54° tan 36°
The above equation can be written as,
cos 2 26 ° + cos (90 - 26)° sin 26° + tan 36° cot (90 - 36)° \text{cos}^2\space26° + \text{cos (90 - 26)° sin 26°} + \dfrac{\text{tan 36°}}{\text{cot (90 - 36)°}} cos 2 26° + cos (90 - 26)° sin 26° + cot (90 - 36)° tan 36°
As, cos(90 - θ) = sin θ and cot(90 - θ) = tan θ. Using in above equation we get,
⇒ cos 2 26 ° + sin 26° sin 26° + tan 36° tan 36° = cos 2 26 ° + sin 2 26 ° + tan 36° tan 36° = 1 + 1 [ ∵ sin 2 A + cos 2 A = 1 ] = 2. \Rightarrow \text{cos}^2\space26° + \text{sin 26° sin 26°} + \dfrac{\text{tan 36°}}{\text{tan 36°}} \\[1em] = \text{cos}^2\space26° + \text{sin}^2\space26° + \dfrac{\text{tan 36°}}{\text{tan 36°}} \\[1em] = 1 + 1 \quad [\because \text{sin}^2\space A + \text{cos}^2\space A = 1] \\[1em] = 2. ⇒ cos 2 26° + sin 26° sin 26° + tan 36° tan 36° = cos 2 26° + sin 2 26° + tan 36° tan 36° = 1 + 1 [ ∵ sin 2 A + cos 2 A = 1 ] = 2.
Hence, the value of the expression is 2.
Without using trigonometric tables, evaluate the following:
sec 17° cosec 73° + tan 68° cot 22° + cos 2 44 ° + cos 2 46 ° . \dfrac{\text{sec 17°}}{\text{cosec 73°}} + \dfrac{\text{tan 68°}}{\text{cot 22°}} + \text{cos}^2\space 44° + \text{cos}^2\space 46°. cosec 73° sec 17° + cot 22° tan 68° + cos 2 44° + cos 2 46°.
Answer
We need to find the value of
sec 17° cosec 73° + tan 68° cot 22° + cos 2 44 ° + cos 2 46 ° \dfrac{\text{sec 17°}}{\text{cosec 73°}} + \dfrac{\text{tan 68°}}{\text{cot 22°}} + \text{cos}^2\space 44° + \text{cos}^2\space 46° cosec 73° sec 17° + cot 22° tan 68° + cos 2 44° + cos 2 46°
The above equation can be written as,
sec 17° cosec (90 - 17)° + tan 68° cot (90 - 68)° + cos 2 ( 90 − 46 ) ° + cos 2 46 ° \dfrac{\text{sec 17°}}{\text{cosec (90 - 17)°}} + \dfrac{\text{tan 68°}}{\text{cot (90 - 68)°}} + \text{cos}^2\space (90 - 46)° + \text{cos}^2\space 46° cosec (90 - 17)° sec 17° + cot (90 - 68)° tan 68° + cos 2 ( 90 − 46 ) ° + cos 2 46°
As, cos(90 - θ) = sin θ, cosec(90 - θ) = sec θ, cot(90 - θ) = tan θ and sin2 θ + cos2 θ = 1. Using in above equation we get,
⇒ sec 17° sec 17° + tan 68° tan 68° + sin 2 46 ° + cos 2 46 ° = 1 + 1 + 1 = 3 \Rightarrow \dfrac{\text{sec 17°}}{\text{sec 17°}} + \dfrac{\text{tan 68°}}{\text{tan 68°}} + \text{sin}^2\space46° + \text{cos}^2\space46° \\[1em] = 1 + 1 + 1 \\[1em] = 3 ⇒ sec 17° sec 17° + tan 68° tan 68° + sin 2 46° + cos 2 46° = 1 + 1 + 1 = 3
Hence, the value of the expression is 3.
Without using trigonometric tables, evaluate the following:
sin 65° cos 25° + cos 32° sin 58° − sin 28° sec 62° + cosec 2 30 ° \dfrac{\text{sin 65°}}{\text{cos 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62°} + \text{cosec}^2 \space 30° cos 25° sin 65° + sin 58° cos 32° − sin 28° sec 62° + cosec 2 30°
Answer
We need to find the value of
sin 65° cos 25° + cos 32° sin 58° − sin 28° sec 62° + cosec 2 30 ° \dfrac{\text{sin 65°}}{\text{cos 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62°} + \text{cosec}^2 \space 30° cos 25° sin 65° + sin 58° cos 32° − sin 28° sec 62° + cosec 2 30°
As sec θ = 1 cos θ \dfrac{1}{\text{cos θ}} cos θ 1
The above equation can be written as,
sin 65° cos (90 - 65)° + cos (90 - 58)° sin 58° − sin 28° 1 cos (90 - 28)° + cosec 2 30 ° \dfrac{\text{sin 65°}}{\text{cos (90 - 65)°}} + \dfrac{\text{cos (90 - 58)°}}{\text{sin 58°}} - \text{sin 28°} \dfrac{1}{\text{cos (90 - 28)°}} + \text{cosec}^2 \space 30° cos (90 - 65)° sin 65° + sin 58° cos (90 - 58)° − sin 28° cos (90 - 28)° 1 + cosec 2 30°
As, cos(90 - θ) = sin θ and cosec 30° = 2. Using in above equation,
= sin 65° sin 65° + sin 58° sin 58° − sin 28° 1 sin 28° + 2 2 = 1 + 1 − 1 + 4 = 6 − 1 = 5. = \dfrac{\text{sin 65°}}{\text{sin 65°}} + \dfrac{\text{sin 58°}}{\text{sin 58°}} - \text{sin 28°} \dfrac{1}{\text{sin 28°}} + 2^2 \\[1em] = 1 + 1 - 1 + 4 \\[1em] = 6 - 1 \\[1em] = 5. = sin 65° sin 65° + sin 58° sin 58° − sin 28° sin 28° 1 + 2 2 = 1 + 1 − 1 + 4 = 6 − 1 = 5.
Hence, the value of the expression is 5.
Without using trigonometric tables, evaluate the following:
sec 29° cosec 61° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° − 3(sin 2 38 ° + sin 2 52 ° ) . \dfrac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3(sin}^2 38° + \text{sin}^2 52°). cosec 61° sec 29° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° − 3(sin 2 38° + sin 2 52° ) .
Answer
We need to find the value of
sec 29° cosec 61° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° − 3(sin 2 38 ° + sin 2 52 ° ) \dfrac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3(sin}^2 38° + \text{sin}^2 52°) cosec 61° sec 29° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° − 3(sin 2 38° + sin 2 52° )
The above equation can be written as,
sec 29° cosec (90 - 29)° + 2 cot 8° cot 17° cot 45° cot (90 - 17)° cot (90 - 8)° − 3(sin 2 ( 90 − 52 ) ° + sin 2 52 ° ) \dfrac{\text{sec 29°}}{\text{cosec (90 - 29)°}} + \text{2 cot 8° cot 17° cot 45° cot (90 - 17)° cot (90 - 8)°} - \text{3(sin}^2 (90 - 52)° + \text{sin}^2 52°) cosec (90 - 29)° sec 29° + 2 cot 8° cot 17° cot 45° cot (90 - 17)° cot (90 - 8)° − 3(sin 2 ( 90 − 52 ) ° + sin 2 52° )
As, sin(90 - θ) = cos θ, cosec(90 - θ) = sec θ, cot(90 - θ) = tan θ, cot θ.tan θ = 1, cot 45° = 1 and sin2 θ + cos2 θ = 1. Using this in above equation we get,
⇒ sec 29° sec 29° + 2 cot 8° cot 17° cot 45° tan 17° tan 8° − 3(cos 2 52 ° + sin 2 52 ° ) = 1 + 2 cot 8° tan 8° cot 45° cot 17° tan 17° − 3(cos 2 52 ° + sin 2 52 ° ) = 1 + 2 × 1 × 1 × 1 − 3 ( 1 ) = 1 + 2 − 3 = 3 − 3 = 0. \Rightarrow \dfrac{\text{sec 29°}}{\text{sec 29°}} + \text{2 cot 8° cot 17° cot 45° tan 17° tan 8°} - \text{3(cos}^2 52° + \text{sin}^2 52°) \\[1em] = 1 + \text{2 cot 8° tan 8° cot 45° cot 17° tan 17°} - \text{3(cos}^2 52° + \text{sin}^2 52°) \\[1em] = 1 + 2 \times 1 \times 1 \times 1 - 3(1) \\[1em] = 1 + 2 - 3 \\[1em] = 3 - 3 \\[1em] = 0. ⇒ sec 29° sec 29° + 2 cot 8° cot 17° cot 45° tan 17° tan 8° − 3(cos 2 52° + sin 2 52° ) = 1 + 2 cot 8° tan 8° cot 45° cot 17° tan 17° − 3(cos 2 52° + sin 2 52° ) = 1 + 2 × 1 × 1 × 1 − 3 ( 1 ) = 1 + 2 − 3 = 3 − 3 = 0.
Hence, the value of the above expression is 0.
Without using trigonometric tables, evaluate the following:
sin 35° cos 55° + cos 35° sin 55° cosec 2 10 ° − tan 2 80 ° \dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°} cosec 2 10° − tan 2 80° sin 35° cos 55° + cos 35° sin 55°
Answer
We need to find the value of
sin 35° cos 55° + cos 35° sin 55° cosec 2 10 ° − tan 2 80 ° \dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°} cosec 2 10° − tan 2 80° sin 35° cos 55° + cos 35° sin 55°
As, sin(90 - θ) = cos θ, cos(90 - θ) = sin θ, tan(90 - θ) = cot θ, sin2 θ + cos2 θ = 1 and cosec2 θ - cot2 θ = 1.
The above equation can be written as,
⇒ sin 35° cos (90 - 35)° + cos 35° sin (90 - 35)° cosec 2 10 ° − tan 2 ( 90 − 10 ) ° = sin 35° sin 35° + cos 35° cos 35° cosec 2 10 ° − cot 2 10 ° = sin 2 35 ° + cos 2 35 ° cosec 2 10 ° − cot 2 10 ° = 1 1 = 1. \Rightarrow\dfrac{\text{sin 35° cos (90 - 35)° + cos 35° sin (90 - 35)°}}{\text{cosec}^2 10° - \text{tan}^2 (90 - 10)°} \\[1em] = \dfrac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{cosec}^2 10° - \text{cot}^2 10°} \\[1em] = \dfrac{\text{sin}^2 35° + \text{cos}^2 35°}{\text{cosec}^2 10° - \text{cot}^2 10°} \\[1em] = \dfrac{1}{1} \\[1em] = 1. ⇒ cosec 2 10° − tan 2 ( 90 − 10 ) ° sin 35° cos (90 - 35)° + cos 35° sin (90 - 35)° = cosec 2 10° − cot 2 10° sin 35° sin 35° + cos 35° cos 35° = cosec 2 10° − cot 2 10° sin 2 35° + cos 2 35° = 1 1 = 1.
Hence, the value of the above expression is 1.
Without using trigonometric tables, evaluate the following:
sin2 34° + sin2 56° + 2 tan 18° tan 72° - cot2 30°.
Answer
We need to find the value of:
sin2 34° + sin2 56° + 2 tan 18° tan 72° - cot2 30°
The above equation can be written as,
sin2 34° + sin2 (90 - 34)° + 2 tan 18° tan (90 - 18)° - cot2 30°.
As, sin(90 - θ) = cos θ, tan(90 - θ) = cot θ, sin2 θ + cos2 θ = 1, tan θ.cot θ = 1 and cot 30° = 3 \sqrt{3} 3
⇒ sin2 34° + cos2 34° + 2 tan 18° cot 18° - 3 2 \sqrt{3}^2 3 2
⇒ 1 + 2 - 3 = 0.
Hence, the value of the above expression is 0.
Without using trigonometric tables, evaluate the following:
( tan 25° cosec 65° ) 2 + ( cot 25° sec 65° ) 2 + 2 tan 18° tan 45° tan 72° . \Big(\dfrac{\text{tan 25°}}{\text{cosec 65°}}\Big)^2 + \Big(\dfrac{\text{cot 25°}}{\text{sec 65°}}\Big)^2 + \text{2 tan 18° tan 45° tan 72°}. ( cosec 65° tan 25° ) 2 + ( sec 65° cot 25° ) 2 + 2 tan 18° tan 45° tan 72° .
Answer
We need to find the value of,
( tan 25° cosec 65° ) 2 + ( cot 25° sec 65° ) 2 + \Big(\dfrac{\text{tan 25°}}{\text{cosec 65°}}\Big)^2 + \Big(\dfrac{\text{cot 25°}}{\text{sec 65°}}\Big)^2 + ( cosec 65° tan 25° ) 2 + ( sec 65° cot 25° ) 2 +
The above equation can be written as,
⇒ [ tan 25° cosec (90 - 25)° ] 2 + [ cot 25° sec (90 - 25)° ] 2 + 2 tan 18 ° × 1 × tan ( 90 − 18 ) ° \Rightarrow \Big[\dfrac{\text{tan 25°}}{\text{cosec (90 - 25)°}}\Big]^2 + \Big[\dfrac{\text{cot 25°}}{\text{sec (90 - 25)°}}\Big]^2 + 2 \text{ tan } 18° \times 1 \times \text{tan } (90 - 18)° \\[1em] ⇒ [ cosec (90 - 25)° tan 25° ] 2 + [ sec (90 - 25)° cot 25° ] 2 + 2 tan 18° × 1 × tan ( 90 − 18 ) °
As, sec(90° - θ) = cosec θ, tan(90° - θ) = cot θ and cosec(90° - θ) = sec θ. Using in above equation we get,
⇒ ( tan 25° sec 25° ) 2 + ( cot 25° cosec 25° ) 2 + 2 tan 18 ° × 1 × cot 18 ° ⇒ ( sin 25° cos 25° 1 cos 25° ) 2 + ( cos 25° sin 25° 1 sin 25° ) 2 + 2 tan 18 ° × 1 tan 18° ⇒ sin 2 25 ° + cos 2 25 ° + 2 ⇒ 1 + 2 = 3. \Rightarrow \Big(\dfrac{\text{tan 25°}}{\text{sec 25°}}\Big)^2 + \Big(\dfrac{\text{cot 25°}}{\text{cosec 25°}}\Big)^2 + 2 \text{ tan } 18° \times 1 \times \text{cot } 18° \\[1em] \Rightarrow \Bigg(\dfrac{\dfrac{\text{sin 25°}}{\text{cos 25°}}}{\dfrac{1}{\text{cos 25°}}}\Bigg)^2 + \Bigg(\dfrac{\dfrac{\text{cos 25°}}{\text{sin 25°}}}{\dfrac{1}{\text{sin 25°}}}\Bigg)^2 + 2\text{ tan } 18° \times \dfrac{1}{\text{tan 18°}} \\[1em] \Rightarrow \text{sin}^2 25° + \text{cos}^2 25° + 2 \\[1em] \Rightarrow 1 + 2 = 3. ⇒ ( sec 25° tan 25° ) 2 + ( cosec 25° cot 25° ) 2 + 2 tan 18° × 1 × cot 18° ⇒ ( cos 25° 1 cos 25° sin 25° ) 2 + ( sin 25° 1 sin 25° cos 25° ) 2 + 2 tan 18° × tan 18° 1 ⇒ sin 2 25° + cos 2 25° + 2 ⇒ 1 + 2 = 3.
Hence, the value of the equation is 3.
Without using trigonometric tables, evaluate the following:
(cos2 25° + cos2 65°) + cosec θ sec(90° - θ) - cot θ tan (90° - θ).
Answer
We need to find the value of,
(cos2 25° + cos2 65°) + cosec θ sec(90° - θ) - cot θ tan (90° - θ).
The above equation can be written as,
⇒ [cos2 25° + cos2 (90 - 25)°] + cosec θ cosec θ - cot θ cot θ
⇒ (cos2 25° + sin2 25°) + cosec θ cosec θ - cot θ cot θ
⇒ 1 + cosec2 θ - cot2 θ
⇒ 1 + 1
⇒ 2.
Hence, the value of the equation is 2.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
(sec A + tan A)(1 - sin A) = cos A.
Answer
The L.H.S of above equation can be written as,
⇒ ( 1 cos A + sin A cos A ) ( 1 − sin A ) ⇒ ( 1 + sin A cos A ) (1 - sin A) ⇒ 1 − sin 2 A cos A ⇒ cos 2 A cos A ⇒ cos A . \Rightarrow \Big(\dfrac{\text{1}}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)(1 - \text{sin A}) \\[1em] \Rightarrow \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big)\text{(1 - sin A)} \\[1em] \Rightarrow \dfrac{1 - \text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A}. ⇒ ( cos A 1 + cos A sin A ) ( 1 − sin A ) ⇒ ( cos A 1 + sin A ) (1 - sin A) ⇒ cos A 1 − sin 2 A ⇒ cos A cos 2 A ⇒ cos A .
Since, L.H.S. = cos A = R.H.S. hence, proved that (sec A + tan A)(1 - sin A) = cos A.
(1 + tan2 A)(1 - sin A)(1 + sin A) = 1.
Answer
The L.H.S of above equation can be written as,
⇒ sec2 A.(1 - sin2 A)
⇒ sec2 A.cos 2 A
⇒ sec 2 A . 1 sec 2 A = 1. \text{sec}^2 A.\dfrac{1}{\text{sec}^2 A} = 1. sec 2 A . sec 2 A 1 = 1.
Since, L.H.S. = 1 = R.H.S. hence, proved that (1 + tan2 A)(1 - sin A)(1 + sin A) = 1.
tan A + cot A = sec A cosec A
Answer
The L.H.S of above equation can be written as,
⇒ sin A cos A + cos A sin A ⇒ sin 2 A + cos 2 A cos A.sin A ⇒ 1 cos A.sin A ⇒ 1 cos A × 1 sin A ⇒ sec A. cosec A . \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A.sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A.sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \times \dfrac{1}{\text{sin A}} \\[1em] \Rightarrow \text{sec A. cosec A}. ⇒ cos A sin A + sin A cos A ⇒ cos A.sin A sin 2 A + cos 2 A ⇒ cos A.sin A 1 ⇒ cos A 1 × sin A 1 ⇒ sec A. cosec A .
Since, L.H.S. = sec A.cosec A = R.H.S., hence proved that tan A + cot A = sec A cosec A.
(1 - cos A)(1 + sec A) = tan A sin A.
Answer
The L.H.S. of the equation can be written as,
⇒ ( 1 − cos A ) ( 1 + 1 cos A ) = ( 1 − cos A ) ( 1 + cos A cos A ) = 1 − cos 2 A cos A = sin 2 A cos A = sin A sin A cos A = tan A sin A . \Rightarrow (1 - \text{cos A})\Big(1 + \dfrac{1}{\text{cos A}}\Big) \\[1em] = (1 - \text{cos A})\Big(\dfrac{1 + \text{cos A}}{\text{cos A}}\Big) \\[1em] = \dfrac{1 - \text{cos}^2 A}{\text{cos A}} \\[1em] = \dfrac{\text{sin}^2 A}{\text{cos A}} \\[1em] = \dfrac{\text{sin A sin A}}{\text{cos A}} \\[1em] = \text{tan A sin A}. ⇒ ( 1 − cos A ) ( 1 + cos A 1 ) = ( 1 − cos A ) ( cos A 1 + cos A ) = cos A 1 − cos 2 A = cos A sin 2 A = cos A sin A sin A = tan A sin A .
Since, L.H.S. = R.H.S. hence proved that (1 - cos A)(1 + sec A) = tan A sin A.
cot2 A - cos2 A = cot2 A cos2 A
Answer
The L.H.S of above equation can be written as,
⇒ cos 2 A sin 2 A − cos 2 A ⇒ cos 2 A − cos 2 A . sin 2 A sin 2 A ⇒ cos 2 A ( 1 − sin 2 A ) sin 2 A ⇒ cos 2 A sin 2 A × cos 2 A ⇒ cot 2 A cos 2 A \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} - \text{cos}^2 A \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2A.\text{sin}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A(1 - \text{sin}^2 A)}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} \times \text{cos}^2 A \\[1em] \Rightarrow \text{cot}^2 A\text{ cos}^2 A ⇒ sin 2 A cos 2 A − cos 2 A ⇒ sin 2 A cos 2 A − cos 2 A . sin 2 A ⇒ sin 2 A cos 2 A ( 1 − sin 2 A ) ⇒ sin 2 A cos 2 A × cos 2 A ⇒ cot 2 A cos 2 A
Since, L.H.S. = cot2 A. cos2 A = R.H.S., hence proved that cot2 A - cos2 A = cot2 A cos2 A.
1 + tan 2 θ 1 + sec θ \dfrac{\text{tan }^2 θ}{1 + \text{sec } θ} 1 + sec θ tan 2 θ
Answer
The L.H.S of above equation can be written as,
⇒ 1 + tan 2 θ 1 + sec θ ⇒ 1 + sec 2 θ − 1 1 + sec θ ⇒ 1 + ( sec θ − 1 ) ( sec θ + 1 ) 1 + sec θ ⇒ 1 + sec θ − 1 ⇒ sec θ . \Rightarrow 1 + \dfrac{\text{tan }^2 θ}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \dfrac{\text{sec }^2 θ - 1}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \dfrac{(\text{sec } θ - 1)(\text{sec } θ + 1)}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \text{sec } θ - 1\\[1em] \Rightarrow \text{sec } θ. ⇒ 1 + 1 + sec θ tan 2 θ ⇒ 1 + 1 + sec θ sec 2 θ − 1 ⇒ 1 + 1 + sec θ ( sec θ − 1 ) ( sec θ + 1 ) ⇒ 1 + sec θ − 1 ⇒ sec θ .
Since, L.H.S. = R.H.S.
Hence, proved 1 + tan 2 θ 1 + sec θ \dfrac{\text{tan }^2 θ}{1 + \text{sec } θ} 1 + sec θ tan 2 θ
1 + sec A sec A = sin 2 A 1 − cos A \dfrac{1 + \text{sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{1 - \text{cos A}} sec A 1 + sec A = 1 − cos A sin 2 A
Answer
The L.H.S of above equation can be written as,
⇒ 1 + 1 cos A 1 cos A ⇒ cos A + 1 cos A 1 cos A ⇒ ( cos A + 1 ) cos A cos A ⇒ cos A + 1 ⇒ ( 1 + cos A ) × 1 − cos A 1 − cos A ⇒ 1 − cos 2 A 1 − cos A ⇒ sin 2 A 1 − cos A . \Rightarrow \dfrac{1 + \dfrac{1}{\text{cos A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos A} + 1}{\text{cos A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{(\text{cos A} + 1)\text{ cos A}}{\text{ cos A}} \\[1em] \Rightarrow \text{cos A} + 1 \\[1em] \Rightarrow (1 + \text{cos A}) \times \dfrac{1 - \text{cos A}}{1 - \text{cos A}} \\[1em] \Rightarrow \dfrac{1 - \text{cos}^2 A}{1 - \text{cos} A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{1 - \text{cos A}}. ⇒ cos A 1 1 + cos A 1 ⇒ cos A 1 cos A cos A + 1 ⇒ cos A ( cos A + 1 ) cos A ⇒ cos A + 1 ⇒ ( 1 + cos A ) × 1 − cos A 1 − cos A ⇒ 1 − cos A 1 − cos 2 A ⇒ 1 − cos A sin 2 A .
Since, L.H.S. = R.H.S. hence, proved that 1 + sec A sec A = sin 2 A 1 − cos A \dfrac{1 + \text{sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{1 - \text{cos A}} sec A 1 + sec A = 1 − cos A sin 2 A
Prove the following identity, where the angles involved are acute angles for which the trigonometric ratios are defined:
sin A 1 - cos A = cosec A + cot A \dfrac{\text{sin A}}{\text{1 - cos A}} = \text{cosec A} + \text{cot A} 1 - cos A sin A = cosec A + cot A
Answer
The R.H.S. of the equation can be written as,
⇒ 1 sin A + cos A sin A ⇒ 1 + cos A sin A ⇒ 1 + cos A sin A × 1 - cos A 1 - cos A ⇒ 1 - cos 2 A sin A ( 1 - cos A ) ⇒ sin 2 A sin A ( 1 - cos A ) [ ∵ sin 2 A + cos 2 A = 1 ] ⇒ sin A 1 - cos A \Rightarrow \dfrac{1}{\text{sin A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + cos A}}{\text{sin A}} \times \dfrac{\text{1 - cos A}}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos}^2 \text{A}}{\text{sin A}(\text{1 - cos A})} \\[1em] \Rightarrow \dfrac{\text{sin}^2 \text{A}}{\text{sin A}(\text{1 - cos A})} \quad [\because \text{sin}^2 \text{A} + \text{cos}^2 \text{A} = 1] \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}} ⇒ sin A 1 + sin A cos A ⇒ sin A 1 + cos A ⇒ sin A 1 + cos A × 1 - cos A 1 - cos A ⇒ sin A ( 1 - cos A ) 1 - cos 2 A ⇒ sin A ( 1 - cos A ) sin 2 A [ ∵ sin 2 A + cos 2 A = 1 ] ⇒ 1 - cos A sin A
Since, RHS = LHS, hence proved that sin A 1 - cos A = cosec A + cot A \dfrac{\text{sin A}}{\text{1 - cos A}} = \text{cosec A} + \text{cot A} 1 - cos A sin A = cosec A + cot A
sin A 1 + cos A = 1 - cos A sin A \dfrac{\text{sin A}}{\text{1 + cos A}} = \dfrac{\text{1 - cos A}}{\text{sin A}} 1 + cos A sin A = sin A 1 - cos A
Answer
The L.H.S of above equation can be written as,
sin A 1 + cos A × 1 − cos A 1 − cos A sin A ( 1 − cos A ) ( 1 + cos A ) ( 1 − cos A ) sin A ( 1 − cos A ) 1 − cos 2 A sin A ( 1 − cos A ) sin 2 A 1 − cos A sin A . \dfrac{\text{sin A}}{\text{1 + cos A}} \times \dfrac{1 - \text{cos A}}{1 - \text{cos A }} \\[1em] \dfrac{\text{sin A}(1 - \text{cos A})}{(1 + \text{cos A})(1 - \text{cos A})} \\[1em] \dfrac{\text{sin A}(1 - \text{cos A})}{1 - \text{cos}^2 A} \\[1em] \dfrac{\text{sin A}(1 - \text{cos A})}{\text{sin}^2 A} \\[1em] \dfrac{1 - \text{cos A}}{\text{sin} A}. 1 + cos A sin A × 1 − cos A 1 − cos A ( 1 + cos A ) ( 1 − cos A ) sin A ( 1 − cos A ) 1 − cos 2 A sin A ( 1 − cos A ) sin 2 A sin A ( 1 − cos A ) sin A 1 − cos A .
Since, L.H.S. = R.H.S., hence proved that sin A 1 + cos A = 1 - cos A sin A \dfrac{\text{sin A}}{\text{1 + cos A}} = \dfrac{\text{1 - cos A}}{\text{sin A}} 1 + cos A sin A = sin A 1 - cos A
1 − tan 2 A cot 2 A − 1 = tan 2 A \dfrac{1 - \text{tan}^2 A}{\text{cot}^2 A - 1} = \text{tan}^2 A cot 2 A − 1 1 − tan 2 A = tan 2 A
Answer
The L.H.S. of the equation can be written as,
⇒ 1 − sin 2 A cos 2 A cos 2 A sin 2 A − 1 ⇒ cos 2 A − sin 2 A cos 2 A cos 2 A − sin 2 A sin 2 A ⇒ cos 2 A − sin 2 A cos 2 A − sin 2 A × sin 2 A cos 2 A ⇒ 1 × tan 2 A ⇒ tan 2 A . \Rightarrow \dfrac{1 - \dfrac{\text{sin}^2 A}{\text{cos}^2 A}}{\dfrac{\text{cos}^2 A}{\text{sin}^2 A} - 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos}^2 A}}{\dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos}^2 A - \text{sin}^2 A} \times \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow 1 \times \text{tan}^2 A \\[1em] \Rightarrow \text{tan}^2 A. ⇒ sin 2 A cos 2 A − 1 1 − cos 2 A sin 2 A ⇒ sin 2 A cos 2 A − sin 2 A cos 2 A cos 2 A − sin 2 A ⇒ cos 2 A − sin 2 A cos 2 A − sin 2 A × cos 2 A sin 2 A ⇒ 1 × tan 2 A ⇒ tan 2 A .
Since, L.H.S. = R.H.S. hence, proved that 1 − tan 2 A cot 2 A − 1 = tan 2 A \dfrac{1 - \text{tan}^2 A}{\text{cot}^2 A - 1} = \text{tan}^2 A cot 2 A − 1 1 − tan 2 A = tan 2 A
sin A 1 + cos A = cosec A - cot A \dfrac{\text{sin A}}{\text{1 + cos A}} = \text{cosec A - cot A} 1 + cos A sin A = cosec A - cot A
Answer
The R.H.S. of the equation can be written as,
⇒ 1 sin A − cos A sin A ⇒ 1 - cos A sin A ⇒ 1 - cos A sin A × 1 + cos A 1 + cos A ⇒ 1 − cos 2 A sin A(1 + cos A) ⇒ sin 2 A sin A(1 + cos A) ⇒ sin A 1 + cos A . \Rightarrow \dfrac{1}{\text{sin A}} - \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A}} \times \dfrac{\text{1 + cos A}}{\text{1 + cos A}} \\[1em] \Rightarrow \dfrac{1 - \text{cos}^2 A}{\text{sin A(1 + cos A)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{sin A(1 + cos A)}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{1 + cos A}}. ⇒ sin A 1 − sin A cos A ⇒ sin A 1 - cos A ⇒ sin A 1 - cos A × 1 + cos A 1 + cos A ⇒ sin A(1 + cos A) 1 − cos 2 A ⇒ sin A(1 + cos A) sin 2 A ⇒ 1 + cos A sin A .
Since, R.H.S. = L.H.S. hence, proved that sin A 1 + cos A = cosec A - cot A \dfrac{\text{sin A}}{\text{1 + cos A}} = \text{cosec A - cot A} 1 + cos A sin A = cosec A - cot A
( 1 − tan θ 1 − cot θ ) 2 \Big(\dfrac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 ( 1 − cot θ 1 − tan θ ) 2 2 θ
Answer
The L.H.S of above equation can be written as,
⇒ ( 1 − tan θ 1 − cot θ ) 2 ⇒ ( 1 − tan θ 1 − 1 tan θ ) 2 ⇒ ( ( 1 − tan θ ) . tan θ tan θ − 1 ) 2 ⇒ ( − ( tan θ − 1 ) . tan θ tan θ − 1 ) 2 ⇒ ( − tan θ ) 2 ⇒ tan 2 θ . \Rightarrow \Big(\dfrac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \text{tan } θ}{1 - \dfrac{1}{\text{tan } θ}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{(1 - \text{tan } θ). \text{tan } θ}{\text{tan } θ - 1} \Big)^2 \\[1em] \Rightarrow \Big(\dfrac{-(\text{tan } θ - 1). \text{tan } θ}{\text{tan } θ - 1} \Big)^2 \\[1em] \Rightarrow (-\text{tan } θ)^2 \\[1em] \Rightarrow \text{tan }^2 θ. ⇒ ( 1 − cot θ 1 − tan θ ) 2 ⇒ ( 1 − tan θ 1 1 − tan θ ) 2 ⇒ ( tan θ − 1 ( 1 − tan θ ) . tan θ ) 2 ⇒ ( tan θ − 1 − ( tan θ − 1 ) . tan θ ) 2 ⇒ ( − tan θ ) 2 ⇒ tan 2 θ .
Since, L.H.S. = R.H.S.
Hence, proved ( 1 − tan θ 1 − cot θ ) 2 \Big(\dfrac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 ( 1 − cot θ 1 − tan θ ) 2 2 θ
sec A - 1 sec A + 1 = 1 - cos A 1 + cos A \dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}} sec A + 1 sec A - 1 = 1 + cos A 1 - cos A
Answer
The L.H.S. of the equation can be written as,
⇒ 1 cos A − 1 1 cos A + 1 ⇒ 1 − cos A cos A 1 + cos A cos A ⇒ (1 - cos A) × cos A (1 + cos A) × cos A ⇒ 1 - cos A 1 + cos A . \Rightarrow \dfrac{{\dfrac{1}{\text{cos A}} - 1}}{\dfrac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 - \text{cos A}}{\text{cos A}}} {\dfrac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A)} \times \text{cos A}}{\text{(1 + cos A)} \times \text{cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{1 + cos A}}. ⇒ cos A 1 + 1 cos A 1 − 1 ⇒ cos A 1 + cos A cos A 1 − cos A ⇒ (1 + cos A) × cos A (1 - cos A) × cos A ⇒ 1 + cos A 1 - cos A .
Since, L.H.S. = R.H.S. hence, proved that sec A - 1 sec A + 1 = 1 - cos A 1 + cos A \dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}} sec A + 1 sec A - 1 = 1 + cos A 1 - cos A
tan 2 θ (sec θ - 1) 2 = 1 + cos θ 1 - cos θ \dfrac{\text{tan}^2 θ}{\text{(sec θ - 1)}^2} = \dfrac{\text{1 + cos θ}}{\text{1 - cos θ}} (sec θ - 1) 2 tan 2 θ = 1 - cos θ 1 + cos θ
Answer
The L.H.S. of the equation can be written as,
⇒ ( sin θ cos θ ) 2 ( 1 cos θ − 1 ) 2 ⇒ sin 2 θ cos 2 θ ( 1 − cos θ cos θ ) 2 \Rightarrow \dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)^2}{\left(\dfrac{1}{\cos\theta}-1\right)^2} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ}}{\left(\dfrac{1-\cos\theta}{\cos\theta}\right)^2} \\[1em] ⇒ ( cos θ 1 − 1 ) 2 ( cos θ sin θ ) 2 ⇒ ( cos θ 1 − cos θ ) 2 cos 2 θ sin 2 θ
Using ( sin 2 θ = 1 − cos 2 θ ) : (\sin^2\theta=1-\cos^2\theta): ( sin 2 θ = 1 − cos 2 θ ) :
⇒ 1 − cos 2 θ cos 2 θ ( 1 − cos θ ) 2 cos 2 θ ⇒ 1 − cos 2 θ ( 1 − cos θ ) 2 ⇒ ( 1 − cos θ ) ( 1 + cos θ ) ( 1 − cos θ ) 2 ⇒ ( 1 − cos θ ) ( 1 + cos θ ) ( 1 − cos θ ) 2 ⇒ 1 + cos θ 1 − cos θ \Rightarrow \dfrac{\dfrac{1-\cos^2\theta}{\cos^2\theta}}{\dfrac{(1-\cos\theta)^2}{\cos^2\theta}} \\[1em] \Rightarrow \dfrac{1-\cos^2\theta}{(1-\cos\theta)^2} \\[1em] \Rightarrow \dfrac{(1-\cos\theta)(1+\cos\theta)}{(1-\cos\theta)^2} \\[1em] \Rightarrow \dfrac{\cancel{(1-\cos\theta)}(1+\cos\theta)}{(1-\cos\theta)^{\cancel{2}}} \\[1em] \Rightarrow \dfrac{1 + \cos\theta}{1 - \cos\theta} \\[1em] ⇒ cos 2 θ ( 1 − cos θ ) 2 cos 2 θ 1 − cos 2 θ ⇒ ( 1 − cos θ ) 2 1 − cos 2 θ ⇒ ( 1 − cos θ ) 2 ( 1 − cos θ ) ( 1 + cos θ ) ⇒ ( 1 − cos θ ) 2 ( 1 − cos θ ) ( 1 + cos θ ) ⇒ 1 − cos θ 1 + cos θ
Since, L.H.S. = R.H.S. hence, proved that tan 2 θ (sec θ - 1) 2 = 1 + cos θ 1 - cos θ \dfrac{\text{tan}^2 θ}{\text{(sec θ - 1)}^2} = \dfrac{\text{1 + cos θ}}{\text{1 - cos θ}} (sec θ - 1) 2 tan 2 θ = 1 - cos θ 1 + cos θ
(1 + tan A)2 + (1 - tan A)2 = 2 sec2 A.
Answer
The L.H.S of the equation can be written as,
⇒ 1 + tan2 A + 2 tan A + 1 + tan2 A - 2 tan A
⇒ 2(1 + tan2 A)
⇒ 2 sec2 A.
Since, L.H.S. = R.H.S hence, proved that (1 + tan A)2 + (1 - tan A)2 = 2 sec2 A.
sec2 A + cosec2 A = sec2 A cosec2 A.
Answer
The L.H.S of the equation can be written as,
⇒ 1 cos 2 A + 1 sin 2 A ⇒ sin 2 A + cos 2 A cos 2 A . sin 2 A ⇒ 1 cos 2 A . sin 2 A ⇒ sec 2 A . cosec 2 A \Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A +\text{cos}^2 A}{\text{cos}^2 A .\text{ sin}^2 A} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A .\text{ sin}^2 A} \\[1em] \Rightarrow \text{sec}^2 A. \text{cosec}^2 A ⇒ cos 2 A 1 + sin 2 A 1 ⇒ cos 2 A . sin 2 A sin 2 A + cos 2 A ⇒ cos 2 A . sin 2 A 1 ⇒ sec 2 A . cosec 2 A
Since, L.H.S. = R.H.S hence, proved that sec2 A + cosec2 A = sec2 A cosec2 A.
1 + sin A cos A + cos A 1 + sin A = 2 sec A \dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A} cos A 1 + sin A + 1 + sin A cos A = 2 sec A
Answer
The L.H.S of the equation can be written as,
⇒ ( 1 + sin A ) 2 + cos 2 A cos A(1 + sin A) ⇒ 1 + sin 2 A + 2 sin A + cos 2 A cos A(1 + sin A) ⇒ 1 + sin 2 A + cos 2 A + 2 sin A cos A(1 + sin A) ⇒ 1 + 1 + 2 sin A cos A(1 + sin A) ⇒ 2 + 2 sin A cos A(1 + sin A) ⇒ 2 ( 1 + sin A ) cos A(1 + sin A) ⇒ 2 cos A ⇒ 2 sec A . \Rightarrow \dfrac{(1 + \text{sin A})^2 + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{1 + \text{sin}^2 A + 2\text{sin A} + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{1 + \text{sin}^2 A + \text{cos}^2 A + 2\text{sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{1 + 1 + 2\text{sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2 + 2\text{sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2(1 + \text{sin A})}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2}{\text{cos A}} \\[1em] \Rightarrow 2\text{ sec A}. ⇒ cos A(1 + sin A) ( 1 + sin A ) 2 + cos 2 A ⇒ cos A(1 + sin A) 1 + sin 2 A + 2 sin A + cos 2 A ⇒ cos A(1 + sin A) 1 + sin 2 A + cos 2 A + 2 sin A ⇒ cos A(1 + sin A) 1 + 1 + 2 sin A ⇒ cos A(1 + sin A) 2 + 2 sin A ⇒ cos A(1 + sin A) 2 ( 1 + sin A ) ⇒ cos A 2 ⇒ 2 sec A .
Since, L.H.S. = R.H.S. hence proved that 1 + sin A cos A + cos A 1 + sin A = 2 sec A \dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A} cos A 1 + sin A + 1 + sin A cos A = 2 sec A
tan A sec A - 1 + tan A sec A + 1 = 2 cosec A \dfrac{\text{tan A}}{\text{sec A - 1}} + \dfrac{\text{tan A}}{\text{sec A + 1}} = 2 \text{ cosec A} sec A - 1 tan A + sec A + 1 tan A = 2 cosec A
Answer
The L.H.S of the equation can be written as,
⇒ tan A ( 1 sec A - 1 + 1 sec A + 1 ) ⇒ tan A [ sec A + 1 + sec A - 1 (sec A - 1)(sec A + 1) ] ⇒ tan A ( 2 sec A sec 2 A − 1 ) ⇒ tan A ( 2 sec A tan 2 A ) ⇒ 2 sec A tan A ⇒ 2 1 cos A sin A cos A ⇒ 2 sin A ⇒ 2 cosec A . \Rightarrow \text{tan A}\Big(\dfrac{1}{\text{sec A - 1}} + \dfrac{1}{\text{sec A + 1}}\Big) \\[1em] \Rightarrow \text{tan A}\Big[\dfrac{\text{sec A + 1 + sec A - 1}}{\text{(sec A - 1)(sec A + 1)}}\Big] \\[1em] \Rightarrow \text{tan A}\Big(\dfrac{\text{2 sec A}}{\text{sec}^2 A - 1}\Big) \\[1em] \Rightarrow \text{tan A}\Big(\dfrac{\text{2 sec A}}{\text{tan}^2 A}\Big) \\[1em] \Rightarrow \dfrac{\text{2 sec A}}{\text{tan A}} \\[1em] \Rightarrow \dfrac{2\dfrac{1}{\text{cos A}}}{\dfrac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{2}{\text{sin A}} \\[1em] \Rightarrow \text{2 cosec A}. ⇒ tan A ( sec A - 1 1 + sec A + 1 1 ) ⇒ tan A [ (sec A - 1)(sec A + 1) sec A + 1 + sec A - 1 ] ⇒ tan A ( sec 2 A − 1 2 sec A ) ⇒ tan A ( tan 2 A 2 sec A ) ⇒ tan A 2 sec A ⇒ cos A sin A 2 cos A 1 ⇒ sin A 2 ⇒ 2 cosec A .
Since, L.H.S. = R.H.S. hence, proved that tan A sec A - 1 + tan A sec A + 1 = 2 cosec A \dfrac{\text{tan A}}{\text{sec A - 1}} + \dfrac{\text{tan A}}{\text{sec A + 1}} = 2 \text{ cosec A} sec A - 1 tan A + sec A + 1 tan A = 2 cosec A
cosec A cosec A - 1 + cosec A cosec A + 1 = 2 sec 2 A \dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}} = 2 \text{ sec}^2 A cosec A - 1 cosec A + cosec A + 1 cosec A = 2 sec 2 A
Answer
The L.H.S of the equation can be written as,
⇒ cosec A ( 1 cosec A - 1 + 1 cosec A + 1 ) ⇒ cosec A [ cosec A + 1 + cosec A - 1 (cosec A - 1)(cosec A + 1) ] ⇒ cosec A ( 2 cosec A cosec 2 A − 1 ) ⇒ cosec A ( 2 cosec A cot 2 A ) ⇒ 2 cosec 2 A cot 2 A ⇒ 2 1 sin 2 A cos 2 A sin 2 A ⇒ 2 cos 2 A ⇒ 2 sec 2 A . \Rightarrow \text{cosec A}\Big(\dfrac{1}{\text{cosec A - 1}} + \dfrac{1}{\text{cosec A + 1}}\Big) \\[1em] \Rightarrow \text{cosec A}\Big[\dfrac{\text{cosec A + 1 + cosec A - 1}}{\text{(cosec A - 1)(cosec A + 1)}}\Big] \\[1em] \Rightarrow \text{cosec A}\Big(\dfrac{\text{2 cosec A}}{\text{cosec}^2 A - 1}\Big) \\[1em] \Rightarrow \text{cosec A}\Big(\dfrac{\text{2 cosec A}}{\text{cot}^2 A}\Big) \\[1em] \Rightarrow \dfrac{\text{2 cosec}^2 A}{\text{cot}^2 A} \\[1em] \Rightarrow \dfrac{2\dfrac{1}{\text{sin}^2 A}}{\dfrac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{2}{\text{cos}^2 A} \\[1em] \Rightarrow \text{2 sec}^2 A. ⇒ cosec A ( cosec A - 1 1 + cosec A + 1 1 ) ⇒ cosec A [ (cosec A - 1)(cosec A + 1) cosec A + 1 + cosec A - 1 ] ⇒ cosec A ( cosec 2 A − 1 2 cosec A ) ⇒ cosec A ( cot 2 A 2 cosec A ) ⇒ cot 2 A 2 cosec 2 A ⇒ sin 2 A cos 2 A 2 sin 2 A 1 ⇒ cos 2 A 2 ⇒ 2 sec 2 A .
Since, L.H.S. = R.H.S. hence, proved that cosec A cosec A - 1 + cosec A cosec A + 1 = 2 sec 2 A \dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}} = 2 \text{ sec}^2 A cosec A - 1 cosec A + cosec A + 1 cosec A = 2 sec 2 A
cot A - tan A = 2 cos 2 A − 1 sin A cos A \text{cot A - tan A} = \dfrac{2\space\text{cos}^2 A - 1}{\text{sin A cos A}} cot A - tan A = sin A cos A 2 cos 2 A − 1
Answer
The L.H.S of the equation can be written as,
⇒ cos A sin A − sin A cos A ⇒ cos 2 A − sin 2 A sin A cos A ⇒ cos 2 A − (1 - cos 2 A ) sin A cos A ⇒ cos 2 A + cos 2 A − 1 sin A cos A ⇒ 2 cos 2 A − 1 sin A cos A \Rightarrow \dfrac{\text{cos A}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{(1 - cos}^2 A)}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + \text{cos}^2 A - 1}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{2\text{cos}^2 A - 1}{\text{sin A cos A}} \\[1em] ⇒ sin A cos A − cos A sin A ⇒ sin A cos A cos 2 A − sin 2 A ⇒ sin A cos A cos 2 A − (1 - cos 2 A ) ⇒ sin A cos A cos 2 A + cos 2 A − 1 ⇒ sin A cos A 2 cos 2 A − 1
Since, L.H.S. = R.H.S. hence, proved that cot A - tan A = 2 cos 2 A − 1 sin A cos A \dfrac{2\text{cos}^2 A - 1}{\text{sin A cos A}} sin A cos A 2 cos 2 A − 1
cot A - 1 2 − sec 2 A = cot A 1 + tan A \dfrac{\text{cot A - 1}}{2 - \text{sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}} 2 − sec 2 A cot A - 1 = 1 + tan A cot A
Answer
The L.H.S of the equation can be written as,
⇒ cos A sin A − 1 2 − 1 cos 2 A ⇒ cos A − sin A sin A 2 cos 2 A − 1 cos 2 A ⇒ cos 2 A ( cos A - sin A ) sin A ( 2 cos 2 A − 1 ) ⇒ cos 2 A ( cos A - sin A ) sin A [ 2 cos 2 A − ( sin 2 A + cos 2 A ) ] ⇒ cos 2 A ( cos A - sin A ) sin A ( 2 cos 2 A − sin 2 A − cos 2 A ) ) ⇒ cos 2 A ( cos A - sin A ) sin A ( 2 cos 2 A − sin 2 A − cos 2 A ) ) ⇒ cos 2 A ( cos A - sin A ) sin A ( cos 2 A − sin 2 A ) ⇒ cos 2 A ( cos A - sin A ) sin A ( cos A - sin A ) ( cos A + sin A ) ⇒ cos 2 A sin A(cos A + sin A) ⇒ cos A . cos A sin A(cos A + sin A) ⇒ cot A. cos A (cos A + sin A) ⇒ cot A. cos A cos A cos A + sin A cos A ⇒ cot A 1 + tan A . \Rightarrow \dfrac{\dfrac{\text{cos A}}{\text{sin A}} - 1}{2 - \dfrac{1}{\text{cos}^2 A}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos A} - \text{sin A}}{\text{sin A}}}{\dfrac{2 \text{cos}^2 A - 1}{\text{cos}^2 A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(2\text{cos}^2 A - 1)} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}[2\text{cos}^2 A - (\text{sin}^2 A + \text{cos}^2 A)]} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(2\text{cos}^2 A - \text{sin}^2 A - \text{cos}^2 A))} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(2\text{cos}^2 A - \text{sin}^2 A - \text{cos}^2 A))} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(\text{cos}^2 A - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A (\text{cos A - sin A})}{\text{sin A}(\text{cos A - sin A}) (\text{cos A + sin A})} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A(cos A + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos A . cos A}}{\text{sin A(cos A + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cot A. cos A}}{\text{(cos A + sin A)}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cot A. cos A}}{\text{cos A}}}{\dfrac{\text{cos A + sin A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{cot A}}{\text{1 + tan A}}. ⇒ 2 − cos 2 A 1 sin A cos A − 1 ⇒ cos 2 A 2 cos 2 A − 1 sin A cos A − sin A ⇒ sin A ( 2 cos 2 A − 1 ) cos 2 A ( cos A - sin A ) ⇒ sin A [ 2 cos 2 A − ( sin 2 A + cos 2 A )] cos 2 A ( cos A - sin A ) ⇒ sin A ( 2 cos 2 A − sin 2 A − cos 2 A )) cos 2 A ( cos A - sin A ) ⇒ sin A ( 2 cos 2 A − sin 2 A − cos 2 A )) cos 2 A ( cos A - sin A ) ⇒ sin A ( cos 2 A − sin 2 A ) cos 2 A ( cos A - sin A ) ⇒ sin A ( cos A - sin A ) ( cos A + sin A ) cos 2 A ( cos A - sin A ) ⇒ sin A(cos A + sin A) cos 2 A ⇒ sin A(cos A + sin A) cos A . cos A ⇒ (cos A + sin A) cot A. cos A ⇒ cos A cos A + sin A cos A cot A. cos A ⇒ 1 + tan A cot A .
Since, L.H.S. = R.H.S. hence, proved that cot A - 1 2 − sec 2 A = cot A 1 + tan A \dfrac{\text{cot A - 1}}{2 - \text{sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}} 2 − sec 2 A cot A - 1 = 1 + tan A cot A
1 1 + sin θ + 1 1 − sin θ \dfrac{1}{1 + \text{sin } θ} + \dfrac{1}{1 - \text{sin } θ} 1 + sin θ 1 + 1 − sin θ 1 2 θ
Answer
The L.H.S of above equation can be written as,
⇒ 1 1 + sin θ + 1 1 − sin θ ⇒ 1 × ( 1 − sin θ ) + 1 × ( 1 + sin θ ) ( 1 + sin θ ) × ( 1 − sin θ ) ⇒ 1 − sin θ + 1 + sin θ 1 2 − ( sin θ ) 2 ⇒ 2 1 − sin 2 θ ⇒ 2 cos 2 θ ⇒ 2 sec 2 θ \Rightarrow \dfrac{1}{1 + \text{sin } θ} + \dfrac{1}{1 - \text{sin } θ} \\[1em] \Rightarrow \dfrac{1 \times (1 - \text{sin } θ) + 1 \times (1 + \text{sin } θ)}{(1 + \text{sin } θ) \times (1 - \text{sin } θ)} \\[1em] \Rightarrow \dfrac{1 - \text{sin } θ + 1 + \text{sin } θ}{1^2 - (\text{sin } θ)^2} \\[1em] \Rightarrow \dfrac{2}{1 - \text{sin }^2 θ} \\[1em] \Rightarrow \dfrac{2}{\text{cos }^2 θ} \\[1em] \Rightarrow 2\text{ sec }^2 θ ⇒ 1 + sin θ 1 + 1 − sin θ 1 ⇒ ( 1 + sin θ ) × ( 1 − sin θ ) 1 × ( 1 − sin θ ) + 1 × ( 1 + sin θ ) ⇒ 1 2 − ( sin θ ) 2 1 − sin θ + 1 + sin θ ⇒ 1 − sin 2 θ 2 ⇒ cos 2 θ 2 ⇒ 2 sec 2 θ
Since, L.H.S. = R.H.S.
Hence, proved 1 1 + sin θ + 1 1 − sin θ \dfrac{1}{1 + \text{sin } θ} + \dfrac{1}{1 - \text{sin } θ} 1 + sin θ 1 + 1 − sin θ 1 2 θ
tan2 θ - sin2 θ = tan2 θ sin2 θ
Answer
The L.H.S of the equation can be written as,
⇒ sin 2 θ cos 2 θ − sin 2 θ ⇒ sin 2 θ − sin 2 θ . cos 2 θ cos 2 θ ⇒ sin 2 θ ( 1 - cos 2 θ ) cos 2 θ ⇒ tan 2 θ . sin 2 θ \Rightarrow \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \text{sin}^2 θ \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ - \text{sin}^2 θ. \text{cos}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ(\text{1 - cos}^2 θ)}{\text{cos}^2 θ} \\[1em] \Rightarrow \text{tan}^2 θ. \text{ sin}^ 2 θ ⇒ cos 2 θ sin 2 θ − sin 2 θ ⇒ cos 2 θ sin 2 θ − sin 2 θ . cos 2 θ ⇒ cos 2 θ sin 2 θ ( 1 - cos 2 θ ) ⇒ tan 2 θ . sin 2 θ
Since, L.H.S. = R.H.S. hence, proved that tan2 θ - sin2 θ = tan2 θ sin2 θ.
cos θ 1 - tan θ − sin 2 θ cos θ - sin θ = cos θ + sin θ \dfrac{\text{cos θ}}{\text{1 - tan θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} = \text{cos θ + sin θ} 1 - tan θ cos θ − cos θ - sin θ sin 2 θ = cos θ + sin θ
Answer
The L.H.S of the equation can be written as,
⇒ cos θ 1 - tan θ − sin 2 θ cos θ - sin θ ⇒ cos θ 1 − sin θ cos θ − sin 2 θ cos θ - sin θ ⇒ cos 2 θ cos θ - sin θ − sin 2 θ cos θ - sin θ ⇒ cos 2 θ − sin 2 θ cos θ - sin θ ⇒ (cos θ - sin θ)(cos θ + sin θ) cos θ - sin θ ⇒ cos θ + sin θ . \Rightarrow \dfrac{\text{cos θ}}{\text{1 - tan θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{cos θ}}{{1 - \dfrac{\text{sin θ}}{\text{cos θ}}}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{cos θ - sin θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{(cos θ - sin θ)(cos θ + sin θ)}}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \text{cos θ + sin θ}. ⇒ 1 - tan θ cos θ − cos θ - sin θ sin 2 θ ⇒ 1 − cos θ sin θ cos θ − cos θ - sin θ sin 2 θ ⇒ cos θ - sin θ cos 2 θ − cos θ - sin θ sin 2 θ ⇒ cos θ - sin θ cos 2 θ − sin 2 θ ⇒ cos θ - sin θ (cos θ - sin θ)(cos θ + sin θ) ⇒ cos θ + sin θ .
Since, L.H.S. = R.H.S. hence, proved that cos θ 1 - tan θ − sin 2 θ cos θ - sin θ = cos θ + sin θ \dfrac{\text{cos θ}}{\text{1 - tan θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} = \text{cos θ + sin θ} 1 - tan θ cos θ − cos θ - sin θ sin 2 θ = cos θ + sin θ
Prove that:
( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 cos 2 θ \dfrac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{ cos }^2θ} 2 cos 2 θ ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 θ + tan2 θ
Answer
The L.H.S of above equation can be written as,
⇒ ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 cos 2 θ ⇒ 1 + sin 2 θ + 2 sin θ + 1 + sin 2 θ − 2 sin θ 2 cos 2 θ ⇒ 2 + 2 sin 2 θ 2 cos 2 θ ⇒ 2 ( 1 + sin 2 θ ) 2 cos 2 θ ⇒ 1 + sin 2 θ cos 2 θ ⇒ 1 cos 2 θ + sin 2 θ cos 2 θ ⇒ sec θ 2 + tan 2 θ \Rightarrow \dfrac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{cos }^2θ} \\[1em] \Rightarrow\dfrac{1 + \text{sin }^2θ + 2\text{sin }θ + 1 + \text{sin }^2θ - 2\text{sin }θ}{2\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{2 + 2\text{sin }^2θ}{2\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{2(1 + \text{sin }^2θ)}{2\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{1 + \text{sin }^2θ}{\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{1}{\text{cos }^2θ} + \dfrac{\text{sin }^2θ}{\text{cos }^2θ}\\[1em] \Rightarrow \text{sec }θ^2 + \text{tan }^2θ\\[1em] ⇒ 2 cos 2 θ ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 ⇒ 2 cos 2 θ 1 + sin 2 θ + 2 sin θ + 1 + sin 2 θ − 2 sin θ ⇒ 2 cos 2 θ 2 + 2 sin 2 θ ⇒ 2 cos 2 θ 2 ( 1 + sin 2 θ ) ⇒ cos 2 θ 1 + sin 2 θ ⇒ cos 2 θ 1 + cos 2 θ sin 2 θ ⇒ sec θ 2 + tan 2 θ
Since, L.H.S. = sec2 θ + tan2 θ = R.H.S.
Hence, proved ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 cos 2 θ \dfrac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{cos }^2θ} 2 cos 2 θ ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 θ + tan2 θ.
cosec4 θ - cosec2 θ = cot4 θ + cot2 θ
Answer
The L.H.S of the equation can be written as,
⇒ cosec2 θ(cosec2 θ - 1)2 θ. cot2 θ2 θ). cot2 θ2 θ + cot4 θ.
Since, L.H.S. = R.H.S. hence, proved that cosec4 θ - cosec2 θ = cot4 θ + cot2 θ.
2 sec2 θ - sec4 θ - 2 cosec2 θ + cosec4 θ = cot4 θ - tan4 θ.
Answer
The L.H.S of the equation can be written as,
⇒ 2(1 + tan2 θ) - (sec2 θ)2 - 2(1 + cot2 θ) + (cosec2 θ)2 2 θ - (1 + tan2 θ)2 - 2 - 2cot2 θ + (1 + cot2 θ)2 2 θ - (1 + tan4 θ + 2tan2 θ) - 2 - 2cot2 θ + (1 + cot4 θ + 2cot2 θ)2 θ - 1 - tan4 θ - 2tan2 θ - 2 - 2cot2 θ + 1 + cot4 θ + 2cot2 θ2 θ - 2tan2 θ + 2cot2 θ - 2cot2 θ + 1 - 1 + cot4 θ - tan4 θ4 θ - tan4 θ.
Since, L.H.S. = R.H.S. hence, proved that 2 sec2 θ - sec4 θ - 2 cosec2 θ + cosec4 θ = cot4 θ - tan4 θ.
1 + cos θ - sin 2 θ sin θ(1 + cos θ) = cot θ \dfrac{\text{1 + cos θ - sin}^2 \text{θ}}{\text{sin θ(1 + cos θ)}} = \text{cot θ} sin θ(1 + cos θ) 1 + cos θ - sin 2 θ = cot θ
Answer
The L.H.S of the equation can be written as,
⇒ 1 + cos θ - (1 - cos 2 θ ) sin θ(1 + cos θ) ⇒ 1 − 1 + cos 2 θ + cos θ sin θ(1 + cos θ) ⇒ cos θ(cos θ + 1) sin θ(1 + cos θ) ⇒ cos θ sin θ ⇒ cot θ . \Rightarrow \dfrac{\text{1 + cos θ - (1 - cos}^2 θ)}{\text{sin θ(1 + cos θ)}} \\[1em] \Rightarrow \dfrac{1 - 1 + \text{cos}^2 θ + \text{cos } θ}{\text{sin θ(1 + cos θ)}} \\[1em] \Rightarrow \dfrac{\text{cos θ(cos θ + 1)}}{\text{sin θ(1 + cos θ)}} \\[1em] \Rightarrow \dfrac{\text{cos θ}}{\text{sin θ}} \\[1em] \Rightarrow \text{cot θ}. ⇒ sin θ(1 + cos θ) 1 + cos θ - (1 - cos 2 θ ) ⇒ sin θ(1 + cos θ) 1 − 1 + cos 2 θ + cos θ ⇒ sin θ(1 + cos θ) cos θ(cos θ + 1) ⇒ sin θ cos θ ⇒ cot θ .
Since, L.H.S. = R.H.S. hence, proved that 1 + cos θ - sin 2 θ sin θ(1 + cos θ) \dfrac{\text{1 + cos θ - sin}^2 θ}{\text{sin θ(1 + cos θ)}} sin θ(1 + cos θ) 1 + cos θ - sin 2 θ
tan 3 θ − 1 tan θ − 1 = sec 2 θ + tan θ . \dfrac{\text{tan}^3 \text{ θ} - 1}{\text{tan θ} - 1} = \text{sec}^2 \text{ θ} + \text{tan θ}. tan θ − 1 tan 3 θ − 1 = sec 2 θ + tan θ .
Answer
As, a3 - b3 = (a - b)(a2 + ab + b2 )
∴ tan3 θ - (1)3 = (tan θ - 1)(tan2 θ + tan θ + 1)
The L.H.S. of the equation can be written as,
⇒ (tan θ - 1) ( tan 2 θ + tan θ + 1 ) tan θ - 1 ⇒ tan 2 θ + tan θ + 1 ⇒ sec 2 θ − 1 + 1 + tan θ ⇒ sec 2 θ + tan θ \Rightarrow \dfrac{\text{(tan θ - 1)}(\text{tan}^2 \text{ θ} + \text{tan θ + 1})}{\text{tan θ - 1}} \\[1em] \Rightarrow \text{tan}^2 \text{ θ} + \text{tan θ + 1} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} - 1 + 1 + \text{tan θ} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} + \text{tan θ} ⇒ tan θ - 1 (tan θ - 1) ( tan 2 θ + tan θ + 1 ) ⇒ tan 2 θ + tan θ + 1 ⇒ sec 2 θ − 1 + 1 + tan θ ⇒ sec 2 θ + tan θ
Since, L.H.S. = R.H.S. hence, proved that tan 3 θ − 1 tan θ - 1 \dfrac{\text{tan}^3 \text{ θ} - 1}{\text{tan θ - 1}} tan θ - 1 tan 3 θ − 1 2 θ + tan θ.
1 + cosec A cosec A = cos 2 A 1 - sin A \dfrac{\text{1 + cosec A}}{\text{cosec A}} = \dfrac{\text{cos}^2 \text{ A}}{\text{1 - sin A}} cosec A 1 + cosec A = 1 - sin A cos 2 A
Answer
The L.H.S. of the equation can be written as,
⇒ 1 + cosec A cosec A ⇒ 1 + 1 sin A 1 sin A ⇒ sin A + 1 sin A 1 sin A ⇒ sin A(sin A + 1) sin A ⇒ sin A + 1 \Rightarrow \dfrac{\text{1 + cosec A}}{\text{cosec A}} \\[1em] \Rightarrow \dfrac{1 + \dfrac{1}{\text{sin A}}}{\dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A + 1}}{\text{sin A}}}{\dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{sin A(sin A + 1)}}{\text{sin A}} \\[1em] \Rightarrow \text{sin A + 1} ⇒ cosec A 1 + cosec A ⇒ sin A 1 1 + sin A 1 ⇒ sin A 1 sin A sin A + 1 ⇒ sin A sin A(sin A + 1) ⇒ sin A + 1
The R.H.S. of the equation can be written as,
⇒ 1 − sin 2 A 1 − sin A ⇒ (1 - sin A)(1 + sin A) 1 - sin A ⇒ 1 + sin A \Rightarrow \dfrac{1 - \text{sin}^2 A}{1 - \text{sin A}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)(1 + sin A)}}{\text{1 - sin A}} \\[1em] \Rightarrow \text{1 + sin A} ⇒ 1 − sin A 1 − sin 2 A ⇒ 1 - sin A (1 - sin A)(1 + sin A) ⇒ 1 + sin A
Since, L.H.S. = R.H.S. hence, proved that 1 + cosec A cosec A = cos 2 A 1 - sin A \dfrac{\text{1 + cosec A}}{\text{cosec A}} = \dfrac{\text{cos}^2 \text{ A}}{\text{1 - sin A}} cosec A 1 + cosec A = 1 - sin A cos 2 A
1 − cos A 1 + cos A = sin A 1 + cos A \sqrt{\dfrac{1 - \text{cos A}}{\text{1 + cos A}}} = \dfrac{\text{sin A}}{\text{1 + cos A}} 1 + cos A 1 − cos A = 1 + cos A sin A
Answer
The L.H.S. of the equation can be written as,
⇒ ( 1 − cos A ) ( 1 + cos A ) (1 + cos A)(1 + cos A) ⇒ ( 1 − cos 2 A ) (1 + cos A) 2 ⇒ sin 2 A (1 + cos A) 2 ⇒ sin A 1 + cos A . \Rightarrow \sqrt{\dfrac{(1 - \text{cos A})(1 + \text{cos A})}{\text{(1 + cos A)(1 + \text{cos A})}}} \\[1em] \Rightarrow \sqrt{\dfrac{(1 - \text{cos}^2 A)}{\text{(1 + cos A)}^2}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{sin}^2 A}{\text{(1 + cos A)}^2}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{1 + cos A}}. ⇒ (1 + cos A)(1 + cos A ) ( 1 − cos A ) ( 1 + cos A ) ⇒ (1 + cos A) 2 ( 1 − cos 2 A ) ⇒ (1 + cos A) 2 sin 2 A ⇒ 1 + cos A sin A .
Since, L.H.S. = R.H.S. hence, proved that 1 − cos A 1 + cos A = sin A 1 + cos A \sqrt{\dfrac{1 - \text{cos A}}{\text{1 + cos A}}} = \dfrac{\text{sin A}}{\text{1 + cos A}} 1 + cos A 1 − cos A = 1 + cos A sin A
1 + sin A 1 - sin A = tan A + sec A \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} = \text{tan A + sec A} 1 - sin A 1 + sin A = tan A + sec A
Answer
The L.H.S. of the equation can be written as,
⇒ (1 + sin A)(1 + sin A) (1 - sin A)(1 + sin A) ⇒ (1 + sin A) 2 (1 - sin 2 A ) ⇒ (1 + sin A) 2 cos 2 A ⇒ 1 + sin A cos A ⇒ 1 cos A + sin A cos A ⇒ sec A + tan A \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)}^2}{\text{(1 - sin}^2 A)}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)}^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A} ⇒ (1 - sin A)(1 + sin A) (1 + sin A)(1 + sin A) ⇒ (1 - sin 2 A ) (1 + sin A) 2 ⇒ cos 2 A (1 + sin A) 2 ⇒ cos A 1 + sin A ⇒ cos A 1 + cos A sin A ⇒ sec A + tan A
Since, L.H.S. = R.H.S. hence, proved that 1 + sin A 1 - sin A \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} 1 - sin A 1 + sin A
1 - cos A 1 + cos A = cosec A - cot A \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = \text{cosec A - cot A} 1 + cos A 1 - cos A = cosec A - cot A
Answer
The L.H.S. of the equation can be written as,
⇒ (1 - cos A)(1 - cos A) (1 + cos A)(1 - cos A) ⇒ (1 - cos A) 2 (1 - cos 2 A ) ⇒ (1 - cos A) 2 sin 2 A ⇒ 1 - cos A sin A ⇒ 1 sin A − cos A sin A ⇒ cosec A - cot A . \Rightarrow \sqrt{\dfrac{\text{(1 - cos A)(1 - cos A)}}{\text{(1 + cos A)(1 - cos A)}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 - cos A)}^2}{\text{(1 - cos}^2 A)}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 - cos A)}^2}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} - \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A - cot A}. ⇒ (1 + cos A)(1 - cos A) (1 - cos A)(1 - cos A) ⇒ (1 - cos 2 A ) (1 - cos A) 2 ⇒ sin 2 A (1 - cos A) 2 ⇒ sin A 1 - cos A ⇒ sin A 1 − sin A cos A ⇒ cosec A - cot A .
Since, L.H.S. = R.H.S. hence, proved that 1 - cos A 1 + cos A \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} 1 + cos A 1 - cos A .
sec A - 1 sec A + 1 + sec A + 1 sec A - 1 = 2 cosec A \sqrt{\dfrac{\text{sec A - 1}}{\text{sec A + 1}}} + \sqrt{\dfrac{\text{sec A + 1}}{\text{sec A - 1}}} = \text{2 cosec A} sec A + 1 sec A - 1 + sec A - 1 sec A + 1 = 2 cosec A
Answer
The L.H.S. of the equation can be written as,
⇒ ( sec A - 1 ) 2 + ( sec A + 1 ) 2 (sec A + 1)(sec A - 1) ⇒ sec A - 1 + sec A + 1 sec 2 A − 1 ⇒ 2 sec A tan 2 A ⇒ 2 sec A tan A ⇒ 2 cos A sin A cos A ⇒ 2 sin A ⇒ 2 cosec A \Rightarrow \dfrac{\sqrt{(\text{sec A - 1})^2} + \sqrt{(\text{sec A + 1})^2}}{\sqrt{\text{(sec A + 1)(sec A - 1)}}} \\[1em] \Rightarrow \dfrac{\text{sec A - 1 + sec A + 1}}{\sqrt{\text{sec}^2 A - 1}} \\[1em] \Rightarrow \dfrac{\text{2 sec A}}{\sqrt{\text{tan}^2 A}} \\[1em] \Rightarrow \dfrac{\text{2 sec A}}{\text{tan A}} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\text{cos A}}}{\dfrac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{2}{\text{sin A}} \\[1em] \Rightarrow 2\text{ cosec A} ⇒ (sec A + 1)(sec A - 1) ( sec A - 1 ) 2 + ( sec A + 1 ) 2 ⇒ sec 2 A − 1 sec A - 1 + sec A + 1 ⇒ tan 2 A 2 sec A ⇒ tan A 2 sec A ⇒ cos A sin A cos A 2 ⇒ sin A 2 ⇒ 2 cosec A
Since, L.H.S. = R.H.S. hence, proved that sec A - 1 sec A + 1 + sec A + 1 sec A - 1 \sqrt{\dfrac{\text{sec A - 1}}{\text{sec A + 1}}} + \sqrt{\dfrac{\text{sec A + 1}}{\text{sec A - 1}}} sec A + 1 sec A - 1 + sec A - 1 sec A + 1
cos A cot A 1 - sin A = 1 + cosec A \dfrac{\text{cos A cot A}}{\text{1 - sin A}} = \text{1 + cosec A} 1 - sin A cos A cot A = 1 + cosec A
Answer
The L.H.S. of the equation can be written as,
⇒ cos A × cos A sin A 1 − sin A ⇒ cos 2 A sin A(1 - sin A) ⇒ 1 - sin 2 A sin A(1 - sin A) ⇒ (1 - sin A)(1 + sin A) sin A(1 - sin A) ⇒ 1 + sin A sin A ⇒ 1 sin A + sin A sin A ⇒ cosec A + 1 \Rightarrow \dfrac{\text{cos A} \times \dfrac{\text{cos A}}{\text{sin A}}}{1 - \text{sin A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A(1 - sin A)}} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A}{\text{sin A(1 - sin A)}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)(1 + sin A)}}{\text{sin A(1 - sin A)}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} + \dfrac{\text{sin A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A + 1} ⇒ 1 − sin A cos A × sin A cos A ⇒ sin A(1 - sin A) cos 2 A ⇒ sin A(1 - sin A) 1 - sin 2 A ⇒ sin A(1 - sin A) (1 - sin A)(1 + sin A) ⇒ sin A 1 + sin A ⇒ sin A 1 + sin A sin A ⇒ cosec A + 1
Since, L.H.S. = R.H.S. hence, proved that cos A cot A 1 - sin A \dfrac{\text{cos A cot A}}{\text{1 - sin A}} 1 - sin A cos A cot A
1 + tan A sin A + 1 + cot A cos A = 2(sec A + cosec A) \dfrac{\text{1 + tan A}}{\text{sin A}} + \dfrac{\text{1 + cot A}}{\text{cos A}} = \text{2(sec A + cosec A)} sin A 1 + tan A + cos A 1 + cot A = 2(sec A + cosec A)
Answer
The L.H.S. of the equation can be written as,
⇒ 1 + sin A cos A sin A + 1 + cos A sin A cos A ⇒ cos A ( 1 + sin A cos A ) + sin A ( 1 + cos A sin A ) sin A cos A ⇒ cos A + sin A + sin A + cos A sin A cos A ⇒ 2(cos A + sin A) sin A cos A ⇒ 2 ( cos A sin A cos A + sin A sin A cos A ) ⇒ 2 ( 1 sin A + 1 cos A ) ⇒ 2 (cosec A + sec A) . \Rightarrow \dfrac{1 + \dfrac{\text{sin A}}{\text{cos A}}}{\text{sin A}} + \dfrac{1 + \dfrac{\text{cos A}}{\text{sin A}}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos A}\Big(1 + \dfrac{\text{sin A}}{\text{cos A}}\Big) + \text{sin A}\Big(1 + \dfrac{\text{cos A}}{\text{sin A}}\Big)}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{cos A + sin A + sin A + cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{sin A cos A}} \\[1em] \Rightarrow 2\Big(\dfrac{\text{cos A}}{\text{sin A cos A}} + \dfrac{\text{sin A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow 2\Big(\dfrac{1}{\text{sin A}} + \dfrac{1}{\text{cos A}}\Big) \\[1em] \Rightarrow 2\text{(cosec A + sec A)}. ⇒ sin A 1 + cos A sin A + cos A 1 + sin A cos A ⇒ sin A cos A cos A ( 1 + cos A sin A ) + sin A ( 1 + sin A cos A ) ⇒ sin A cos A cos A + sin A + sin A + cos A ⇒ sin A cos A 2(cos A + sin A) ⇒ 2 ( sin A cos A cos A + sin A cos A sin A ) ⇒ 2 ( sin A 1 + cos A 1 ) ⇒ 2 (cosec A + sec A) .
Since, L.H.S. = R.H.S. hence, proved that 1 + tan A sin A + 1 + cot A cos A \dfrac{\text{1 + tan A}}{\text{sin A}} + \dfrac{\text{1 + cot A}}{\text{cos A}} sin A 1 + tan A + cos A 1 + cot A
sin A 1 + cot A − cos A 1 + tan A \dfrac{\text{sin }A}{1 + \text{cot }A} - \dfrac{\text{cos }A}{1 + \text{tan }A} 1 + cot A sin A − 1 + tan A cos A
Answer
The L.H.S of above equation can be written as,
⇒ sin A ( 1 + cot A ) − cos A ( 1 + tan A ) ⇒ sin A ( 1 + cos A sin A ) − cos A ( 1 + sin A cos A ) ⇒ sin A ( sin A + cos A sin A ) − cos A ( cos A + sin A cos A ) ⇒ sin A × sin A sin A + cos A − cos A × cos A cos A + sin A ⇒ sin 2 A − cos 2 A sin A + cos A ⇒ ( sin A − cos A ) ( sin A + cos A ) sin A + cos A ⇒ sin A − cos A . \Rightarrow \dfrac{\text{sin }A}{(1 + \text{cot }A)} - \dfrac{\text{cos }A}{(1 + \text{tan }A)} \\[1em] \Rightarrow \dfrac{\text{sin }A}{\Big(1 + \dfrac{\text{cos }A}{\text{sin }A}\Big)} - \dfrac{\text{cos }A}{\Big(1 + \dfrac{\text{sin }A}{\text{cos }A}\Big)} \\[1em] \Rightarrow \dfrac{\text{sin }A}{\Big(\dfrac{\text{sin }A + \text{cos }A}{\text{sin }A}\Big)} - \dfrac{\text{cos }A}{\Big(\dfrac{\text{cos }A + \text{sin }A}{\text{cos }A} \Big)} \\[1em] \Rightarrow \dfrac{\text{sin }A \times \text{sin }A}{\text{sin }A + \text{cos }A} - \dfrac{\text{cos }A \times \text{cos }A}{\text{cos }A + \text{sin }A} \\[1em] \Rightarrow \dfrac{\text{sin }^2A - \text{cos }^2A}{\text{sin }A + \text{cos }A}\\[1em] \Rightarrow \dfrac{(\text{sin }A - \text{cos }A)(\text{sin }A + \text{cos }A)}{\text{sin }A + \text{cos }A}\\[1em] \Rightarrow \text{sin }A - \text{cos }A. ⇒ ( 1 + cot A ) sin A − ( 1 + tan A ) cos A ⇒ ( 1 + sin A cos A ) sin A − ( 1 + cos A sin A ) cos A ⇒ ( sin A sin A + cos A ) sin A − ( cos A cos A + sin A ) cos A ⇒ sin A + cos A sin A × sin A − cos A + sin A cos A × cos A ⇒ sin A + cos A sin 2 A − cos 2 A ⇒ sin A + cos A ( sin A − cos A ) ( sin A + cos A ) ⇒ sin A − cos A .
Since, L.H.S. = R.H.S.
Hence, proved sin A ( 1 + cot A ) − cos A ( 1 + tan A ) \dfrac{\text{sin }A}{(1 + \text{cot }A)} - \dfrac{\text{cos }A}{(1 + \text{tan }A)} ( 1 + cot A ) sin A − ( 1 + tan A ) cos A
sec4 A - tan4 A = 1 + 2 tan2 A
Answer
The L.H.S. of the equation can be written as,
⇒ (sec2 A - tan2 A)(sec2 A + tan2 A)
⇒ 1 × (sec2 A + tan2 A)
⇒ (sec2 A + tan2 A)
⇒ (1 + tan2 A + tan2 A)
⇒ 1 + 2 tan2 A
Since, L.H.S. = R.H.S. hence, proved that sec4 A - tan4 A = 1 + 2tan2 A.
cosec6 A - cot6 A = 3cot2 A cosec2 A + 1.
Answer
a3 - b3 = (a - b)3 + 3ab(a - b)
∴ L.H.S. of the equation can be written as,
⇒ cosec6 A - cot6 A = (cosec2 A - cot2 A )3 + 3cosec2 A cot2 A(cosec2 A - cot2 A)
⇒ 13 + 3cosec2 A cot2 A × 12 A cot2 A
Since, L.H.S. = R.H.S. hence, proved that cosec6 A - cot6 A = 3cot2 A cosec2 A + 1.
sec6 A - tan6 A = 1 + 3 tan2 A + 3 tan4 A.
Answer
a3 - b3 = (a - b)3 + 3ab(a - b)
∴ L.H.S. of the equation can be written as,
⇒ sec6 A - tan6 A = (sec2 A - tan2 A)3 + 3sec2 A tan2 A(sec2 A - tan2 A)
⇒ 13 + 3sec2 A tan2 A × 1
⇒ 1 + 3sec2 A tan2 A
⇒ 1 + 3(1 + tan2 A)tan2 A
⇒ 1 + 3(tan4 A + tan2 A)
⇒ 1 + 3 tan2 A + 3 tan4 A
Since, L.H.S. = R.H.S. hence, proved that sec6 A - tan6 A = 1 + 3tan2 A + 3tan4 A.
cot θ + cosec θ - 1 cot θ - cosec θ + 1 = 1 + cos θ sin θ \dfrac{\text{cot θ + cosec θ - 1}}{\text{cot θ - cosec θ + 1}} = \dfrac{\text{1 + cos θ}}{\text{sin θ}} cot θ - cosec θ + 1 cot θ + cosec θ - 1 = sin θ 1 + cos θ
Answer
L.H.S. of the equation can be written as,
⇒ cos θ sin θ + 1 sin θ − 1 cos θ sin θ − 1 sin θ + 1 ⇒ cos θ + 1 - sin θ sin θ cos θ - 1 + sin θ sin θ ⇒ cos θ + 1 - sin θ cos θ - 1 + sin θ ⇒ cos θ + (1 - sin θ) cos θ - (1 - sin θ) ⇒ cos θ + (1 - sin θ) cos θ - (1 - sin θ) × cos θ + (1 - sin θ) cos θ + (1 - sin θ) ⇒ [ cos θ + (1 - sin θ) ] 2 cos 2 θ − ( 1 - sin θ ) 2 ⇒ cos 2 θ + (1 - sin θ) 2 + 2 cos θ(1 - sin θ) cos 2 θ − ( 1 + sin 2 θ − 2sin θ ) ⇒ cos 2 θ + sin 2 θ + 1 + 2cos θ - 2sin θ - 2 sin θ cos θ cos 2 θ − 1 − sin 2 θ + 2sin θ ⇒ 1 + 1 + 2cos θ - 2sin θ - 2 sin θ cos θ 1 − sin 2 θ − 1 − sin 2 θ + 2 sin θ ⇒ 2 + 2cos θ - 2sin θ - 2sin θ cos θ 2sin θ - 2sin 2 θ ⇒ 2(1 + cos θ) - 2sin θ(1 + cos θ) 2sin θ(1 - sin θ) ⇒ (1 + cos θ)(2 - 2sin θ) 2sin θ(1 - sin θ) ⇒ 2(1 + cos θ)(1 - sin θ) 2sin θ(1 - sin θ) ⇒ 1 + cos θ sin θ . ] \Rightarrow \dfrac{\dfrac{\text{cos θ}}{\text{sin θ}} + \dfrac{1}{\text{sin θ}} - 1}{\dfrac{\text{cos θ}}{\text{sin θ}} - \dfrac{1}{\text{sin θ}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos θ + 1 - sin θ}}{\text{sin θ}}}{\dfrac{\text{cos θ - 1 + sin θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{{\text{cos θ + 1 - sin θ}}}{\text{cos θ - 1 + sin θ}} \\[1em] \Rightarrow \dfrac{{\text{cos θ + (1 - sin θ)}}}{\text{cos θ - (1 - sin θ)}} \\[1em] \Rightarrow \dfrac{{\text{cos θ + (1 - sin θ)}}}{\text{cos θ - (1 - sin θ)}} \times \dfrac{{\text{cos θ + (1 - sin θ)}}}{{\text{cos θ + (1 - sin θ)}}} \\[1em] \Rightarrow \dfrac{[\text{cos θ + (1 - sin θ)}]^2}{\text{cos}^2 θ - (\text{1 - sin θ})^2} \\[1em] \Rightarrow \dfrac{\text{cos}^2 \text{ θ} + \text{(1 - sin θ)}^2 + 2\text{cos θ(1 - sin θ)}}{\text{cos}^2 \text{ θ} - (1 + \text{sin}^2 θ - \text{2sin θ})} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ + \text{sin}^2 θ + 1 + \text{2cos θ - 2sin θ - 2 sin θ cos θ}}{\text{cos}^2 \text{ θ} - 1 - \text{sin}^2 θ + \text{2sin θ}} \\[1em] \Rightarrow \dfrac{1 + 1 + \text{2cos θ - 2sin θ - 2 sin θ cos θ}}{1 - \text{sin}^2 θ - 1 - \text{sin}^2 θ + \text{2 sin θ}} \\[1em] \Rightarrow \dfrac{2 + \text{2cos θ - 2sin θ - 2sin θ cos θ}}{\text{2sin θ - 2sin}^2 θ} \\[1em] \Rightarrow \dfrac{\text{2(1 + cos θ) - 2sin θ(1 + cos θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{(1 + cos θ)(2 - 2sin θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{2(1 + cos θ)(1 - sin θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{1 + cos θ}}{\text{sin θ}}.] ⇒ sin θ cos θ − sin θ 1 + 1 sin θ cos θ + sin θ 1 − 1 ⇒ sin θ cos θ - 1 + sin θ sin θ cos θ + 1 - sin θ ⇒ cos θ - 1 + sin θ cos θ + 1 - sin θ ⇒ cos θ - (1 - sin θ) cos θ + (1 - sin θ) ⇒ cos θ - (1 - sin θ) cos θ + (1 - sin θ) × cos θ + (1 - sin θ) cos θ + (1 - sin θ) ⇒ cos 2 θ − ( 1 - sin θ ) 2 [ cos θ + (1 - sin θ) ] 2 ⇒ cos 2 θ − ( 1 + sin 2 θ − 2sin θ ) cos 2 θ + (1 - sin θ) 2 + 2 cos θ(1 - sin θ) ⇒ cos 2 θ − 1 − sin 2 θ + 2sin θ cos 2 θ + sin 2 θ + 1 + 2cos θ - 2sin θ - 2 sin θ cos θ ⇒ 1 − sin 2 θ − 1 − sin 2 θ + 2 sin θ 1 + 1 + 2cos θ - 2sin θ - 2 sin θ cos θ ⇒ 2sin θ - 2sin 2 θ 2 + 2cos θ - 2sin θ - 2sin θ cos θ ⇒ 2sin θ(1 - sin θ) 2(1 + cos θ) - 2sin θ(1 + cos θ) ⇒ 2sin θ(1 - sin θ) (1 + cos θ)(2 - 2sin θ) ⇒ 2sin θ(1 - sin θ) 2(1 + cos θ)(1 - sin θ) ⇒ sin θ 1 + cos θ . ]
Since, L.H.S. = R.H.S. hence proved that cot θ + cosec θ - 1 cot θ - cosec θ + 1 = 1 + cos θ sin θ \dfrac{\text{cot θ + cosec θ - 1}}{\text{cot θ - cosec θ + 1}} = \dfrac{\text{1 + cos θ}}{\text{sin θ}} cot θ - cosec θ + 1 cot θ + cosec θ - 1 = sin θ 1 + cos θ
sin θ cot θ + cosec θ = 2 + sin θ cot θ - cosec θ \dfrac{\text{sin θ}}{\text{cot θ + cosec θ}} = 2 + \dfrac{\text{sin θ}}{\text{cot θ - cosec θ}} cot θ + cosec θ sin θ = 2 + cot θ - cosec θ sin θ
Answer
L.H.S. of the equation can be written as,
⇒ sin θ cos θ + 1 sin θ ⇒ sin 2 θ 1 + cos θ ⇒ 1 - cos 2 θ 1 + cos θ ⇒ (1 - cos θ)(1 + cos θ) 1 + cos θ ⇒ 1 - cos θ . \Rightarrow \dfrac{\text{sin θ}}{\dfrac{\text{cos θ + 1}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ}{\text{1 + cos θ}} \\[1em] \Rightarrow \dfrac{\text{1 - cos}^2 θ}{\text{1 + cos θ}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos θ)(1 + cos θ)}}{\text{1 + cos θ}} \\[1em] \Rightarrow \text{1 - cos θ}. ⇒ sin θ cos θ + 1 sin θ ⇒ 1 + cos θ sin 2 θ ⇒ 1 + cos θ 1 - cos 2 θ ⇒ 1 + cos θ (1 - cos θ)(1 + cos θ) ⇒ 1 - cos θ .
R.H.S. of the equation can be written as,
⇒ 2 + sin θ cos θ sin θ − 1 sin θ ⇒ 2 + sin 2 θ cos θ - 1 ⇒ 2(cos θ - 1) + sin 2 θ cos θ - 1 ⇒ 2(cos θ - 1) + 1 - cos 2 θ cos θ - 1 ⇒ 2(cos θ - 1) + (1 + cos θ)(1 - cos θ) cos θ - 1 ⇒ 2(cos θ - 1) - (cos θ - 1)(1 + cos θ) cos θ - 1 ⇒ (cos θ - 1)[2 - (1 + cos θ)] cos θ - 1 ⇒ 2 - 1 - cos θ ⇒ 1 - cos θ \Rightarrow 2 + \dfrac{\text{sin θ}}{\dfrac{\text{cos θ}}{\text{sin θ}} - \dfrac{1}{\text{sin θ}}} \\[1em] \Rightarrow 2 + \dfrac{\text{sin}^2 θ}{\text{cos θ - 1}} \\[1em] \Rightarrow \dfrac{\text{2(cos θ - 1)} + \text{sin}^2 θ}{\text{cos θ - 1}} \\[1em] \Rightarrow \dfrac{\text{2(cos θ - 1)} + \text{1 - cos}^2 θ}{\text{cos θ - 1}} \\[1em] \Rightarrow \dfrac{\text{2(cos θ - 1) + (1 + cos θ)(1 - cos θ)}}{\text{cos θ - 1}} \\[1em] \Rightarrow \dfrac{\text{2(cos θ - 1) - (cos θ - 1)(1 + cos θ)}}{\text{cos θ - 1}} \\[1em] \Rightarrow \dfrac{\text{(cos θ - 1)[2 - (1 + cos θ)]}}{\text{cos θ - 1}} \\[1em] \Rightarrow \text{2 - 1 - cos θ} \\[1em] \Rightarrow \text{1 - cos θ} ⇒ 2 + sin θ cos θ − sin θ 1 sin θ ⇒ 2 + cos θ - 1 sin 2 θ ⇒ cos θ - 1 2(cos θ - 1) + sin 2 θ ⇒ cos θ - 1 2(cos θ - 1) + 1 - cos 2 θ ⇒ cos θ - 1 2(cos θ - 1) + (1 + cos θ)(1 - cos θ) ⇒ cos θ - 1 2(cos θ - 1) - (cos θ - 1)(1 + cos θ) ⇒ cos θ - 1 (cos θ - 1)[2 - (1 + cos θ)] ⇒ 2 - 1 - cos θ ⇒ 1 - cos θ
Since, L.H.S. = 1 - cos θ = R.H.S. hence proved that,
sin θ cot θ + cosec θ = 2 + sin θ cot θ - cosec θ \dfrac{\text{sin θ}}{\text{cot θ + cosec θ}} = 2 + \dfrac{\text{sin θ}}{\text{cot θ - cosec θ}} cot θ + cosec θ sin θ = 2 + cot θ - cosec θ sin θ
(sin θ + cos θ)(sec θ + cosec θ) = 2 + sec θ cosec θ.
Answer
L.H.S. of the equation can be written as,
⇒ (sin θ + cos θ) ( 1 cos θ + 1 sin θ ) ⇒ (sin θ + cos θ)(sin θ + cos θ) sin θ cos θ ⇒ sin 2 θ + cos 2 θ + 2 sin θ cos θ sin θ cos θ ⇒ 1 + 2 sin θ cos θ sin θ cos θ ⇒ 1 sin θ cos θ + 2 sin θ cos θ sin θ cos θ ⇒ cosec θ sec θ + 2. \Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{1}{\text{cos θ}} + \dfrac{1}{\text{sin θ}}\Big) \\[1em] \Rightarrow \dfrac{\text{(sin θ + cos θ)(sin θ + cos θ)}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ + \text{cos}^2 θ + \text{2 sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{1 + \text{2 sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{1}{\text{sin θ cos θ}} + \dfrac{\text{2 sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \text{cosec θ sec θ} + 2. ⇒ (sin θ + cos θ) ( cos θ 1 + sin θ 1 ) ⇒ sin θ cos θ (sin θ + cos θ)(sin θ + cos θ) ⇒ sin θ cos θ sin 2 θ + cos 2 θ + 2 sin θ cos θ ⇒ sin θ cos θ 1 + 2 sin θ cos θ ⇒ sin θ cos θ 1 + sin θ cos θ 2 sin θ cos θ ⇒ cosec θ sec θ + 2.
Since, L.H.S. = R.H.S. hence proved that, (sin θ + cos θ)(sec θ + cosec θ) = 2 + sec θ cosec θ.
(cosec A - sin A)(sec A - cos A)sec2 A = tan A.
Answer
L.H.S. of the equation can be written as,
⇒ ( 1 sin A − sin A ) ( 1 cos A − cos A ) × 1 cos 2 A ⇒ ( 1 - sin 2 A sin A ) ( 1 - cos 2 A cos A ) × 1 cos 2 A \Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big)\Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big) \times \dfrac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin}^2 A}{\text{sin A}}\Big)\Big(\dfrac{\text{1 - cos}^2 A}{\text{cos A}}\Big) \times \dfrac{1}{\text{cos}^2 A} \\[1em] ⇒ ( sin A 1 − sin A ) ( cos A 1 − cos A ) × cos 2 A 1 ⇒ ( sin A 1 - sin 2 A ) ( cos A 1 - cos 2 A ) × cos 2 A 1
As 1 - sin2 A = cos2 A and 1 - cos2 A = sin2 A.
⇒ cos 2 A sin 2 A sin A cos A × 1 cos 2 A ⇒ cos 2 A sin 2 A cos 3 A sin A \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 A}{\text{sin A cos A}} \times \dfrac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 A}{\text{cos}^3 A \text{ sin} A} \\[1em] ⇒ sin A cos A cos 2 A sin 2 A × cos 2 A 1 ⇒ cos 3 A sin A cos 2 A sin 2 A
Dividing numerator and denominator by sin A cos A.
⇒ sin A cos A cos 2 A ⇒ sin A cos A ⇒ tan A \Rightarrow \dfrac{\text{sin A cos A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A} ⇒ cos 2 A sin A cos A ⇒ cos A sin A ⇒ tan A
Since, L.H.S. = R.H.S. hence proved that, (cosec A - sin A)(sec A - cos A)sec2 A = tan A.
sin 3 A + cos 3 A sin A + cos A + sin 3 A − cos 3 A sin A - cos A = 2 \dfrac{\text{sin}^3 A + \text{cos}^3 A}{\text{sin A + cos A}} + \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin A - cos A}} = 2 sin A + cos A sin 3 A + cos 3 A + sin A - cos A sin 3 A − cos 3 A = 2
Answer
We know that,
a3 + b3 = (a + b)(a2 - ab + b2 ).
and
a3 - b3 = (a - b)(a2 + ab + b2 ).
Using above formulas, the L.H.S. of the equation can be written as,
⇒ (sin A + cos A)(sin 2 A − sin A cos A + cos 2 A ) sin A + cos A + (sin A - cos A)(sin 2 A + sin A cos A + cos 2 A ) sin A - cos A ⇒ (sin A + cos A) (sin 2 A − sin A cos A + cos 2 A ) (sin A + cos A) + (sin A - cos A) (sin 2 A + sin A cos A + cos 2 A ) (sin A - cos A) ⇒ sin 2 A + cos 2 A − sin A cos A + sin 2 A + cos 2 A + sin A cos A ⇒ 1 + 1 ⇒ 2. \Rightarrow \dfrac{\text{(sin A + cos A)(sin}^2 A - \text{sin A cos A + cos}^2 A)}{\text{sin A + cos A}} + \dfrac{\text{(sin A - cos A)(sin}^2 A + \text{sin A cos A + cos}^2 A)}{\text{sin A - cos A}} \\[1em] \Rightarrow \dfrac{\cancel{\text{(sin A + cos A)}}\text{(sin}^2 A - \text{sin A cos A + cos}^2 A)}{\cancel{\text{(sin A + cos A)}}} + \dfrac{\cancel{\text{(sin A - cos A)}} \text{(sin}^2 A + \text{sin A cos A + cos}^2 A)}{\cancel{\text{(sin A - cos A)}}} \\[1em] \Rightarrow \text{sin}^2 A + \text{cos}^2 A - \text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A + \text{sin A cos A} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ sin A + cos A (sin A + cos A)(sin 2 A − sin A cos A + cos 2 A ) + sin A - cos A (sin A - cos A)(sin 2 A + sin A cos A + cos 2 A ) ⇒ (sin A + cos A) (sin A + cos A) (sin 2 A − sin A cos A + cos 2 A ) + (sin A - cos A) (sin A - cos A) (sin 2 A + sin A cos A + cos 2 A ) ⇒ sin 2 A + cos 2 A − sin A cos A + sin 2 A + cos 2 A + sin A cos A ⇒ 1 + 1 ⇒ 2.
Since, L.H.S. = R.H.S. hence proved that, sin 3 A + cos 3 A sin A + cos A + sin 3 A − cos 3 A sin A - cos A = 2 \dfrac{\text{sin}^3 A + \text{cos}^3 A}{\text{sin A + cos A}} + \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin A - cos A}} = 2 sin A + cos A sin 3 A + cos 3 A + sin A - cos A sin 3 A − cos 3 A = 2
tan 2 A 1 + tan 2 A + cot 2 A 1 + cot 2 A = 1. \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\text{cot}^2 A}{\text{1 + cot}^2 A} = 1. 1 + tan 2 A tan 2 A + 1 + cot 2 A cot 2 A = 1.
Answer
The L.H.S. of the equation can be written as,
⇒ tan 2 A 1 + tan 2 A + 1 tan 2 A 1 + 1 tan 2 A = tan 2 A 1 + tan 2 A + 1 tan 2 A tan 2 A + 1 tan 2 A = tan 2 A 1 + tan 2 A + 1 tan 2 A + 1 = tan 2 A + 1 tan 2 A + 1 1. \Rightarrow \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\dfrac{1}{\text{tan}^2 A}}{1 + \dfrac{1}{\text{tan}^2 A}} \\[1em] = \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\dfrac{1}{\text{tan}^2 A}}{\dfrac{\text{tan}^2 A + 1}{\text{tan}^2 A}} \\[1em] = \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{1}{\text{tan}^2 A + 1} \\[1em] = \dfrac{\text{tan}^2 A + 1}{\text{tan}^2 A + 1} \\[1em] 1. ⇒ 1 + tan 2 A tan 2 A + 1 + tan 2 A 1 tan 2 A 1 = 1 + tan 2 A tan 2 A + tan 2 A tan 2 A + 1 tan 2 A 1 = 1 + tan 2 A tan 2 A + tan 2 A + 1 1 = tan 2 A + 1 tan 2 A + 1 1.
Since, L.H.S. = R.H.S. hence proved that, tan 2 A 1 + tan 2 A + cot 2 A 1 + cot 2 A = 1. \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\text{cot}^2 A}{\text{1 + cot}^2 A} = 1. 1 + tan 2 A tan 2 A + 1 + cot 2 A cot 2 A = 1.
1 sec A + tan A − 1 cos A = 1 cos A − 1 sec A - tan A \dfrac{1}{\text{sec A + tan A}} - \dfrac{1}{\text{cos A}} = \dfrac{1}{\text{cos A}} - \dfrac{1}{\text{sec A - tan A}} sec A + tan A 1 − cos A 1 = cos A 1 − sec A - tan A 1
Answer
The equation can be written as,
1 sec A + tan A + 1 sec A - tan A = 2 cos A \dfrac{1}{\text{sec A + tan A}} + \dfrac{1}{\text{sec A - tan A}} = \dfrac{2}{\text{cos A}} sec A + tan A 1 + sec A - tan A 1 = cos A 2
The L.H.S. of the equation can be written as,
⇒ sec A - tan A + sec A + tan A (sec A - tan A)(sec A + tan A) = 2 sec A sec 2 A − tan 2 A = 2 sec A = 2 cos A . \Rightarrow \dfrac{\text{sec A - tan A + sec A + tan A}}{\text{(sec A - tan A)(sec A + tan A)}} \\[1em] = \dfrac{\text{2 sec A}}{\text{sec}^2 A - \text{tan}^2 A} \\[1em] = \text{2 sec A} \\[1em] = \dfrac{2}{\text{cos A}}. ⇒ (sec A - tan A)(sec A + tan A) sec A - tan A + sec A + tan A = sec 2 A − tan 2 A 2 sec A = 2 sec A = cos A 2 .
Since, L.H.S. = R.H.S. hence proved that,
1 sec A + tan A − 1 cos A = 1 cos A − 1 sec A - tan A \dfrac{1}{\text{sec A + tan A}} - \dfrac{1}{\text{cos A}} = \dfrac{1}{\text{cos A}} - \dfrac{1}{\text{sec A - tan A}} sec A + tan A 1 − cos A 1 = cos A 1 − sec A - tan A 1
(sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2 .
Answer
The L.H.S. of the equation can be written as,
⇒ sin 2 A + sec 2 A + 2 sin A sec A + cos 2 A + cosec 2 A + 2 cos A cosec A = sin 2 A + cos 2 A + sec 2 A + cosec 2 A + 2 sin A cos A + 2 cos A sin A = 1 + 1 cos 2 A + 1 sin 2 A + 2sin 2 A + 2cos 2 A sin A cos A = 1 + sin 2 A + cos 2 A sin 2 A cos 2 A + 2 ( sin 2 A + cos 2 A ) sin A cos A = 1 + 1 sin 2 A cos 2 A + 2 sin A cos A = ( 1 + 1 sin A cos A ) 2 = ( 1 + sec A cosec A ) 2 . \Rightarrow \text{sin}^2 A + \text{sec}^2 A + \text{2 sin A sec A} + \text{cos}^2 A + \text{cosec}^2 A + \text{2 cos A cosec A} \\[1em] = \text{sin}^2 A + \text{cos}^2 A + \text{sec}^2 A + \text{cosec}^2 A + \dfrac{\text{2 sin A}}{\text{cos A}} + \dfrac{\text{2 cos A}}{\text{sin A}} \\[1em] = 1 + \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A} + \dfrac{\text{2sin}^2 A + \text{2cos}^2 A}{\text{sin A cos A}} \\[1em] = 1 + \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A \text{ cos}^2 A} + \dfrac{2(\text{sin}^2 A + \text{cos}^2 A)}{\text{sin A cos A}} \\[1em] = 1 + \dfrac{1}{\text{sin}^2 A \text{ cos}^2 A} + \dfrac{2}{{\text{sin A cos A}}} \\[1em] = \Big(1 + \dfrac{1}{\text{sin A cos A}}\Big)^2 \\[1em] = \Big(\text{1 + sec A cosec A}\Big)^2. ⇒ sin 2 A + sec 2 A + 2 sin A sec A + cos 2 A + cosec 2 A + 2 cos A cosec A = sin 2 A + cos 2 A + sec 2 A + cosec 2 A + cos A 2 sin A + sin A 2 cos A = 1 + cos 2 A 1 + sin 2 A 1 + sin A cos A 2sin 2 A + 2cos 2 A = 1 + sin 2 A cos 2 A sin 2 A + cos 2 A + sin A cos A 2 ( sin 2 A + cos 2 A ) = 1 + sin 2 A cos 2 A 1 + sin A cos A 2 = ( 1 + sin A cos A 1 ) 2 = ( 1 + sec A cosec A ) 2 .
Since, L.H.S. = R.H.S. hence proved that, (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2 .
tan A + sin A tan A - sin A = sec A + 1 sec A - 1 . \dfrac{\text{tan A + sin A}}{\text{tan A - sin A}} = \dfrac{\text{sec A + 1}}{\text{sec A - 1}}. tan A - sin A tan A + sin A = sec A - 1 sec A + 1 .
Answer
The L.H.S. of the equation can be written as,
⇒ sin A cos A + sin A sin A cos A − sin A = sin A ( 1 cos A + 1 ) sin A ( 1 cos A − 1 ) = ( 1 cos A + 1 ) ( 1 cos A − 1 ) = sec A + 1 sec A - 1 \Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}} + \text{sin A}}{\dfrac{\text{sin A}}{\text{cos A}} - \text{sin A}} \\[1em] = \dfrac{\text{sin A}\Big(\dfrac{1}{\text{cos A}} + 1\Big)}{\text{sin A}\Big(\dfrac{1}{\text{cos A}} - 1\Big)} \\[1em] = \dfrac{\Big(\dfrac{1}{\text{cos A}} + 1\Big)}{\Big(\dfrac{1}{\text{cos A}} - 1\Big)} \\[1em] = \dfrac{\text{sec A + 1}}{\text{sec A - 1}} ⇒ cos A sin A − sin A cos A sin A + sin A = sin A ( cos A 1 − 1 ) sin A ( cos A 1 + 1 ) = ( cos A 1 − 1 ) ( cos A 1 + 1 ) = sec A - 1 sec A + 1
Since, L.H.S. = R.H.S. hence proved that, tan A + sin A tan A - sin A = sec A + 1 sec A - 1 . \dfrac{\text{tan A + sin A}}{\text{tan A - sin A}} = \dfrac{\text{sec A + 1}}{\text{sec A - 1}}. tan A - sin A tan A + sin A = sec A - 1 sec A + 1 .
If sin θ + cos θ = 2 \sqrt{2} 2 2 \sqrt{2} 2
Answer
Given,
sin θ + cos θ = 2 \sqrt{2} 2 2 \sqrt{2} 2
Dividing both sides by sin θ
⇒ 1 + cot θ = 2 \sqrt{2} 2 2 \sqrt{2} 2 2 \sqrt{2} 2 1 2 − 1 \dfrac{1}{\sqrt{2} - 1} 2 − 1 1
Rationalizing,
⇒ cot θ = 1 2 − 1 × 2 + 1 2 + 1 ⇒ cot θ = 2 + 1 2 − 1 = 2 + 1. ⇒ \text{cot θ} = \dfrac{1}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[1em] ⇒ \text{cot θ} = \dfrac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1. ⇒ cot θ = 2 − 1 1 × 2 + 1 2 + 1 ⇒ cot θ = 2 − 1 2 + 1 = 2 + 1.
Hence, proved that cot θ = 2 \sqrt{2} 2
If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.
Answer
Given,
⇒ 7 sin2 θ + 3 cos2 θ = 4
⇒ 4 sin2 θ + 3 sin2 θ + 3 cos2 θ = 4
⇒ 4 sin2 θ + 3(sin2 θ + cos2 θ) = 4
⇒ 4 sin2 θ + 3 = 4
⇒ 4 sin2 θ = 4 - 3
⇒ 4 sin2 θ = 1
⇒ sin2 θ = 1 4 \dfrac{1}{4} 4 1
Taking square root of both the sides we get,
⇒ sin θ = 1 4 \sqrt{\dfrac{1}{4}} 4 1
⇒ sin θ = 1 2 \dfrac{1}{2} 2 1
∴ θ = 30°.
Hence, the value of θ = 30°.
If sec θ + tan θ = m and sec θ - tan θ = n, prove that mn = 1.
Answer
mn = (sec θ + tan θ)(sec θ - tan θ)
mn = (sec2 θ - tan2 θ)
By, trigonometric identities sec2 θ - tan2 θ = 1.
∴ mn = 1.
Hence, proved that mn = 1.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 - y2 = a2 - b2 .
Answer
Given,
x = a sec θ + b tan θ and y = a tan θ + b sec θ.
∴ x2 - y2 = (a sec θ + b tan θ)2 - (tan θ + b sec θ)2
⇒ x2 - y2 = a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ - (a2 tan2 θ + b2 sec 2 θ + 2ab sec θ tan θ)
⇒ x2 - y2 = a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ - a2 tan2 θ - b2 sec 2 θ - 2ab sec θ tan θ
⇒ x2 - y2 = a2 (sec2 θ - tan2 θ) - b2 (sec2 θ - tan2 θ)
⇒ x2 - y2 = a2 - b2 .
Hence, proved that x2 - y2 = a2 - b2 .
If x = h + a cos θ and y = k + a sin θ, prove that (x - h)2 + (y - k)2 = a2 .
Answer
Given, x = h + a cos θ and y = k + a sin θ
∴ (x - h)2 + (y - k)2 = (h + a cos θ - h)2 + (k + a sin θ - k)2 2 + (y - k)2 = (a cos θ)2 + (a sin θ)2 2 + (y - k)2 = a2 cos2 θ + a2 sin2 θ2 + (y - k)2 = a2 (cos2 θ + sin2 θ)2 + (y - k)2 = a2 .
Hence, proved that (x - h)2 + (y - k)2 = a2 .
If θ is an acute angle and cosec θ = 5 \sqrt{5} 5
Answer
sin θ = 1 cosec θ = 1 5 \dfrac{1}{\text{cosec θ}} = \dfrac{1}{\sqrt{5}} cosec θ 1 = 5 1
cos2 θ = 1 - sin2 θ = 1 - ( 1 5 ) 2 = 1 − 1 5 = 4 5 . \Big(\dfrac{1}{\sqrt{5}}\Big)^2 = 1 - \dfrac{1}{5} = \dfrac{4}{5}. ( 5 1 ) 2 = 1 − 5 1 = 5 4 .
cos θ = 4 5 \sqrt{\dfrac{4}{5}} 5 4 2 5 \dfrac{2}{\sqrt{5}} 5 2
cot θ = cos θ sin θ = 2 5 1 5 \dfrac{\text{cos θ}}{\text{sin θ}} = \dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{\sqrt{5}}} sin θ cos θ = 5 1 5 2
cot θ - cos θ = 2 − 2 5 = 2 ( 1 − 1 5 ) = 2 ( 5 − 1 ) 5 . \text{cot θ - cos θ} = 2 - \dfrac{2}{\sqrt{5}} \\[1em] = 2\Big(1 - \dfrac{1}{\sqrt{5}}\Big) \\[1em] = \dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}. cot θ - cos θ = 2 − 5 2 = 2 ( 1 − 5 1 ) = 5 2 ( 5 − 1 ) .
Hence, the value of cot θ - cos θ = 2 ( 5 − 1 ) 5 \dfrac{2(\sqrt{5} - 1)}{\sqrt{5}} 5 2 ( 5 − 1 )
If θ is an acute angle and tan θ = 8 15 \dfrac{8}{15} 15 8
Answer
sec2 θ = 1 + tan2 θ
sec2 θ = 1 + ( 8 15 ) 2 \Big(\dfrac{8}{15}\Big)^2 ( 15 8 ) 2
sec2 θ = 1 + 64 225 = 225 + 64 225 = 289 225 \dfrac{64}{225} = \dfrac{225 + 64}{225} = \dfrac{289}{225} 225 64 = 225 225 + 64 = 225 289
sec θ = 289 225 = 17 15 \sqrt{\dfrac{289}{225}} = \dfrac{17}{15} 225 289 = 15 17
cot θ = 1 tan θ = 1 8 15 = 15 8 . \dfrac{1}{\text{tan θ}} = \dfrac{1}{\dfrac{8}{15}} = \dfrac{15}{8}. tan θ 1 = 15 8 1 = 8 15 .
cosec2 θ = 1 + cot2 θ
cosec2 θ = 1 + ( 15 8 ) 2 \Big(\dfrac{15}{8}\Big)^2 ( 8 15 ) 2
cosec2 θ = 1 + 225 64 = 64 + 225 64 = 289 64 \dfrac{225}{64} = \dfrac{64 + 225}{64} = \dfrac{289}{64} 64 225 = 64 64 + 225 = 64 289
cosec θ = 289 64 = 17 8 \sqrt{\dfrac{289}{64}} = \dfrac{17}{8} 64 289 = 8 17
sec θ + cosec θ = 17 15 + 17 8 = 17 × 8 + 17 × 15 120 = 136 + 255 120 = 391 120 = 3 31 120 . \text{sec θ + cosec θ} = \dfrac{17}{15} + \dfrac{17}{8} \\[1em] = \dfrac{17 \times 8 + 17 \times 15}{120} \\[1em] = \dfrac{136 + 255}{120} \\[1em] = \dfrac{391}{120} \\[1em] = 3\dfrac{31}{120}. sec θ + cosec θ = 15 17 + 8 17 = 120 17 × 8 + 17 × 15 = 120 136 + 255 = 120 391 = 3 120 31 .
Hence, the value of expression sec θ + cosec θ = 3 31 120 . 3\dfrac{31}{120}. 3 120 31 .
Evaluate the following :
2 × ( cos 2 20 ° + cos 2 70 ° sin 2 25 ° + sin 2 65 ° ) 2 \times \Big(\dfrac{\text{cos}^2 20° + \text{cos}^2 70°}{\text{sin}^2 25° + \text{sin}^2 65°}\Big) 2 × ( sin 2 25° + sin 2 65° cos 2 20° + cos 2 70° )
Answer
Since, angles are acute in the equation,
∴ cos(90° - θ) = sin θ, sin(90° - θ) = cos θ and tan(90° - θ) = cot θ.
Using above equations in
2 x ( cos 2 20 ° + cos 2 70 ° sin 2 25 ° + sin 2 65 ° ) \Big(\dfrac{\text{cos}^2 20° + \text{cos}^2 70°}{\text{sin}^2 25° + \text{sin}^2 65°}\Big) ( sin 2 25° + sin 2 65° cos 2 20° + cos 2 70° )
= 2 x ( cos 2 20 ° + cos 2 ( 90 − 20 ) ° sin 2 25 ° + sin 2 ( 90 − 25 ) ° ) \Big(\dfrac{\text{cos}^2 20° + \text{cos}^2 (90 - 20)°}{\text{sin}^2 25° + \text{sin}^2 (90 - 25)°}\Big) ( sin 2 25° + sin 2 ( 90 − 25 ) ° cos 2 20° + cos 2 ( 90 − 20 ) ° )
= 2 x ( cos 2 20 ° + sin 2 20 ° sin 2 25 ° + cos 2 25 ° ) \Big(\dfrac{\text{cos}^2 20° + \text{sin}^2 20°}{\text{sin}^2 25° + \text{cos}^2 25°}\Big) ( sin 2 25° + cos 2 25° cos 2 20° + sin 2 20° )
= 2 x 1 - 1 + tan 13° cot 13° tan 23° cot 23° tan 30°
= 1 + 1 x 1 3 \dfrac{1}{\sqrt{3}} 3 1
= 3 + 1 3 \dfrac{\sqrt{3} + 1}{\sqrt{3}} 3 3 + 1
= ( 3 + 1 ) 3 3 × 3 \dfrac{(\sqrt{3} + 1)\sqrt{3}}{\sqrt{3} \times \sqrt{3}} 3 × 3 ( 3 + 1 ) 3
= 3 + 3 3 \dfrac{3 + \sqrt{3}}{\sqrt{3}} 3 3 + 3
Hence, the value of the expression is 3 + 3 3 \dfrac{3 + \sqrt{3}}{\sqrt{3}} 3 3 + 3
Evaluate the following :
sin 2 22 ° + sin 2 68 ° cos 2 22 ° + cos 2 68 ° \dfrac{\text{sin}^2 22° + \text{sin}^2 68°}{\text{cos}^2 22° + \text{cos}^2 68°} cos 2 22° + cos 2 68° sin 2 22° + sin 2 68° 2 63° + cos 63° sin 27°
Answer
Since, angles are acute in the equation,
∴ cos(90° - θ) = sin θ, sin(90° - θ) = cos θ.
Using above equations in
sin 2 22 ° + sin 2 68 ° cos 2 22 ° + cos 2 68 ° \dfrac{\text{sin}^2 22° + \text{sin}^2 68°}{\text{cos}^2 22° + \text{cos}^2 68°} cos 2 22° + cos 2 68° sin 2 22° + sin 2 68° 2 63° + cos 63° sin 27°
= sin 2 22 ° + sin 2 ( 90 − 22 ) ° cos 2 22 ° + cos 2 ( 90 − 22 ) ° \dfrac{\text{sin}^2 22° + \text{sin}^2 (90 - 22)°}{\text{cos}^2 22° + \text{cos}^2 (90 - 22)°} cos 2 22° + cos 2 ( 90 − 22 ) ° sin 2 22° + sin 2 ( 90 − 22 ) ° 2 63° +cos 63° sin (90 - 63)°
= sin 2 22 ° + cos 2 22 ° cos 2 22 ° + sin 2 22 ° \dfrac{\text{sin}^2 22° + \text{cos}^2 22°}{\text{cos}^2 22° + \text{sin}^2 22°} cos 2 22° + sin 2 22° sin 2 22° + cos 2 22° 2 63° + cos 63° cos 63°
= 1 1 \dfrac{1}{1} 1 1 2 63° + cos2 63°
= 1 + 1
= 2.
Hence, the value of expression is 2.
If 4 3 \dfrac{4}{3} 3 4 2 59° - cot2 31°) - 2 3 \dfrac{2}{3} 3 2 2 56° tan2 34° = x 3 \dfrac{x}{3} 3 x
Answer
Solving the L.H.S of above equation using trigonometric identities,
⇒ 4 3 \dfrac{4}{3} 3 4 2 59° - cot2 (90 - 59)°] - 2 3 \dfrac{2}{3} 3 2 2 56° tan2 (90 - 56)°
= 4 3 \dfrac{4}{3} 3 4 2 59° - tan2 59°) - 2 3 \dfrac{2}{3} 3 2 2 56° cot2 56°
= 4 3 \dfrac{4}{3} 3 4 2 3 \dfrac{2}{3} 3 2
= 4 3 \dfrac{4}{3} 3 4 2 3 \dfrac{2}{3} 3 2
= 2 3 \dfrac{2}{3} 3 2
= 11 3 \dfrac{11}{3} 3 11
Comparing it with R.H.S i.e.,
11 3 = x 3 \dfrac{11}{3} = \dfrac{x}{3} 3 11 = 3 x
x = 11.
Hence, the value of x = 11.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
cos A 1 - sin A + cos A 1 + sin A = 2 sec A . \dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = \text{2 sec A}. 1 - sin A cos A + 1 + sin A cos A = 2 sec A .
Answer
Solving L.H.S.,
⇒ cos A(1 + sin A) + cos A(1 - sin A) (1 - sin A)(1 + sin A) = cos A + cos A sin A + cos A - cos A sin A 1 − sin 2 A = 2 cos A cos 2 A = 2 cos A = 2 sec A . \Rightarrow \dfrac{\text{cos A(1 + sin A) + \text{cos A(1 - sin A)}}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \dfrac{\text{cos A + cos A sin A + cos A - cos A sin A}}{1 - \text{sin}^2 A} \\[1em] = \dfrac{2\text{ cos A}}{\text{cos}^2 A} \\[1em] = \dfrac{2}{\text{cos A}} \\[1em] = 2\text{ sec A}. ⇒ (1 - sin A)(1 + sin A) cos A(1 + sin A) + cos A(1 - sin A) = 1 − sin 2 A cos A + cos A sin A + cos A - cos A sin A = cos 2 A 2 cos A = cos A 2 = 2 sec A .
Since, L.H.S. = R.H.S. hence, proved that cos A 1 - sin A + cos A 1 + sin A = 2 sec A \dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A} 1 - sin A cos A + 1 + sin A cos A = 2 sec A
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
cos A cosec A + 1 + cos A cosec A - 1 = 2 tan A \dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = \text{2 tan A} cosec A + 1 cos A + cosec A - 1 cos A = 2 tan A
Answer
Solving L.H.S.,
⇒ cos A(cosec A - 1) + cos A(cosec A + 1) (cosec A - 1)(cosec A + 1) = cos A cosec A - cos A + cos A cosec A + cos A cosec 2 A − 1 = 2 cos A × 1 sin A cot 2 A = 2 cot A cot 2 A = 2 cot A = 2 tan A . \Rightarrow \dfrac{\text{cos A(cosec A - 1) + \text{cos A(cosec A + 1)}}}{\text{(cosec A - 1)(cosec A + 1)}} \\[1em] = \dfrac{\text{cos A cosec A - cos A + cos A cosec A + cos A}}{\text{cosec}^2 A - 1} \\[1em] = \dfrac{2\text{ cos A } \times \dfrac{1}{\text{sin A}}}{\text{cot}^2 A} \\[1em] = \dfrac{2\text{cot A}}{\text{cot}^2 A} \\[1em] = \dfrac{2}{\text{cot A}} \\[1em] = 2 \text{ tan A}. ⇒ (cosec A - 1)(cosec A + 1) cos A(cosec A - 1) + cos A(cosec A + 1) = cosec 2 A − 1 cos A cosec A - cos A + cos A cosec A + cos A = cot 2 A 2 cos A × sin A 1 = cot 2 A 2 cot A = cot A 2 = 2 tan A .
Since, L.H.S. = R.H.S. hence, proved that cos A cosec A + 1 + cos A cosec A - 1 = 2 tan A \dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = 2 \text{ tan A} cosec A + 1 cos A + cosec A - 1 cos A = 2 tan A
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
(cos θ - sin θ)(1 + tan θ) 2 cos 2 θ − 1 = sec θ . \dfrac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{cos}^2 \text{ θ} - 1} = \text{sec θ}. 2 cos 2 θ − 1 (cos θ - sin θ)(1 + tan θ) = sec θ .
Answer
Solving L.H.S.,
⇒ (cos θ - sin θ) ( 1 + sin θ cos θ ) 2 cos 2 θ − 1 = (cos θ - sin θ)(cos θ + sin θ) cos θ 2 cos 2 θ − 1 = cos 2 θ − sin 2 θ cos θ ( 2 cos 2 θ − 1 ) = cos 2 θ − ( 1 − cos 2 θ ) cos θ(2 cos 2 θ − 1 ) = cos 2 θ + cos 2 θ − 1 cos θ(2 cos 2 θ − 1 ) = (2 cos 2 θ − 1 ) cos θ(2 cos 2 θ − 1 ) = 1 cos θ = sec θ . \Rightarrow \dfrac{\text{(cos θ - sin θ)}(1 + \dfrac{\text{sin θ}}{\text{cos θ}})}{\text{2 cos}^2 \text{ θ} - 1} \\[1em] = \dfrac{\dfrac{\text{(cos θ - sin θ)(cos θ + sin θ)}}{\text{cos θ}}}{\text{2 cos}^2 \text{ θ} - 1} \\[1em] = \dfrac{\text{cos}^2 \text{ θ} - \text{sin}^2 θ}{\text{cos θ}(\text{2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{\text{cos}^2 \text{ θ} - (1 - \text{cos}^2 \text{ θ})}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{\text{cos}^2 \text{ θ} + \text{cos}^2 \text{ θ} - 1}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{\text{(2 cos}^2 \text{ θ} - 1)}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{1}{\text{cos θ}} \\[1em] = \text{sec θ}. ⇒ 2 cos 2 θ − 1 (cos θ - sin θ) ( 1 + cos θ sin θ ) = 2 cos 2 θ − 1 cos θ (cos θ - sin θ)(cos θ + sin θ) = cos θ ( 2 cos 2 θ − 1 ) cos 2 θ − sin 2 θ = cos θ(2 cos 2 θ − 1 ) cos 2 θ − ( 1 − cos 2 θ ) = cos θ(2 cos 2 θ − 1 ) cos 2 θ + cos 2 θ − 1 = cos θ(2 cos 2 θ − 1 ) (2 cos 2 θ − 1 ) = cos θ 1 = sec θ .
Since, L.H.S. = R.H.S. hence, proved that (cos θ - sin θ)(1 + tan θ) 2 cos 2 θ = sec θ \dfrac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{ cos}^2 \text{ θ}} = \text{sec θ} 2 cos 2 θ (cos θ - sin θ)(1 + tan θ) = sec θ
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Answer
Solving L.H.S.,
⇒ sin2 θ + cos4 θ
= 1 - cos2 θ + (cos2 θ)2
= 1 - cos2 θ + (1 - sin2 θ)2
= 1 - cos2 θ + 1 + sin4 θ - 2sin2 θ
= 1 - cos2 θ + 1 + sin4 θ - 2(1 - cos2 θ)
= 2 - cos2 θ + sin4 θ - 2 + 2cos2 θ
= 2cos2 θ - cos2 θ + sin4 θ
= cos2 θ + sin4 θ.
Since, L.H.S. = R.H.S. hence, proved that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
cot θ cosec θ + 1 + cosec θ + 1 cot θ = 2 sec θ . \dfrac{\text{cot θ}}{\text{cosec θ + 1}} + \dfrac{\text{cosec θ + 1}}{\text{cot θ}} = \text{2 sec θ}. cosec θ + 1 cot θ + cot θ cosec θ + 1 = 2 sec θ .
Answer
Solving L.H.S.,
⇒ cot 2 θ + (cosec θ + 1) 2 (cot θ)(cosec θ + 1) = cot 2 θ + cosec 2 θ + 1 + 2 cosec θ (cot θ)(cosec θ + 1) = cot 2 θ + 1 + cosec 2 θ + 2 cosec θ (cot θ)(cosec θ + 1) = cosec 2 θ + cosec 2 θ + 2 cosec θ (cot θ)(cosec θ + 1) = 2 cosec 2 θ + 2 cosec θ (cot θ)(cosec θ + 1) = 2 cosec θ(cosec θ + 1) (cot θ)(cosec θ + 1) = 2 cosec θ cot θ = 2 sin θ cos θ sin θ = 2 cos θ = 2 sec θ . \Rightarrow \dfrac{\text{cot}^2 \text{ θ} + \text{(cosec θ + 1)}^2}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{cot}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 1 + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{cot}^2 \text{ θ} + 1 + \text{cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{cosec}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{2\text{ cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{2 cosec θ(cosec θ + 1)}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{2 cosec θ}}{\text{cot θ}} \\[1em] = \dfrac{\dfrac{2}{\text{sin θ}}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] = \dfrac{2}{\text{cos θ}} \\[1em] = 2\text{ sec θ}. ⇒ (cot θ)(cosec θ + 1) cot 2 θ + (cosec θ + 1) 2 = (cot θ)(cosec θ + 1) cot 2 θ + cosec 2 θ + 1 + 2 cosec θ = (cot θ)(cosec θ + 1) cot 2 θ + 1 + cosec 2 θ + 2 cosec θ = (cot θ)(cosec θ + 1) cosec 2 θ + cosec 2 θ + 2 cosec θ = (cot θ)(cosec θ + 1) 2 cosec 2 θ + 2 cosec θ = (cot θ)(cosec θ + 1) 2 cosec θ(cosec θ + 1) = cot θ 2 cosec θ = sin θ cos θ sin θ 2 = cos θ 2 = 2 sec θ .
Since, L.H.S. = R.H.S hence proved that cot θ cosec θ + 1 + cosec θ + 1 cot θ = 2 sec θ \dfrac{\text{cot θ}}{\text{cosec θ + 1}} + \dfrac{\text{cosec θ + 1}}{\text{cot θ}} = 2\text{ sec θ} cosec θ + 1 cot θ + cot θ cosec θ + 1 = 2 sec θ
Prove the following trigonometry identity :
(sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ
Answer
Solving,
⇒ (sin θ + cos θ)(cosec θ - sec θ) ⇒ (sin θ + cos θ) × ( 1 sin θ − 1 cos θ ) ⇒ (sin θ + cos θ) × ( cos θ - sin θ sin θ cos θ ) ⇒ cos 2 θ − sin 2 θ sin θ. cos θ ⇒ 1 - 2 sin 2 θ sin θ.cos θ [ ∵ cos 2 θ = 1 − sin 2 θ ] ⇒ 1 sin θ.cos θ − 2 sin 2 θ sin θ.cos θ ⇒ cosec θ.sec θ − 2 sin 2 θ sin θ.cos θ ⇒ cosec θ.sec θ - 2 tan θ . \phantom{\Rightarrow} \text{(sin θ + cos θ)(cosec θ - sec θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\dfrac{1}{\text{sin θ}} - \dfrac{1}{\text{cos θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\dfrac{\text{cos θ - sin θ}}{\text{sin θ cos θ}}\Big) \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{\text{1 - 2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \quad [\because \text{cos}^2 \text{ θ} = 1 - \text{sin}^2 \text{ θ}] \\[1em] \Rightarrow \dfrac{1}{\text{sin θ.cos θ}} - \dfrac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ} - \dfrac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ - 2 tan θ}. ⇒ (sin θ + cos θ)(cosec θ - sec θ) ⇒ (sin θ + cos θ) × ( sin θ 1 − cos θ 1 ) ⇒ (sin θ + cos θ) × ( sin θ cos θ cos θ - sin θ ) ⇒ sin θ. cos θ cos 2 θ − sin 2 θ ⇒ sin θ.cos θ 1 - 2 sin 2 θ [ ∵ cos 2 θ = 1 − sin 2 θ ] ⇒ sin θ.cos θ 1 − sin θ.cos θ 2 sin 2 θ ⇒ cosec θ.sec θ − sin θ.cos θ 2 sin 2 θ ⇒ cosec θ.sec θ - 2 tan θ .
Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
sec4 A(1 - sin4 A) - 2 tan2 A = 1.
Answer
Solving L.H.S.,
⇒ 1 cos 4 A (1 + sin 2 A ) (1 - sin 2 A ) − 2 sin 2 A cos 2 A = (1 + sin 2 A ) cos 2 A cos 4 A − 2 sin 2 A cos 2 A = 1 + sin 2 A cos 2 A − 2 sin 2 A cos 2 A = 1 + sin 2 A − 2 sin 2 A cos 2 A = 1 − sin 2 A cos 2 A = cos 2 A cos 2 A = 1. \Rightarrow \dfrac{1}{\text{cos}^4 A}\text{(1 + sin }^2 A)\text{(1 - sin}^2 A) - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{\text{(1 + sin}^2 A)\text{cos}^2 A}{{\text{cos}^4 A}} - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 + \text{sin}^2 A}{\text{cos}^2 A} - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 + \text{sin}^2 A - 2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 - \text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{\text{cos}^2 A}{\text{cos}^2 A} \\[1em] = 1. ⇒ cos 4 A 1 (1 + sin 2 A ) (1 - sin 2 A ) − cos 2 A 2 sin 2 A = cos 4 A (1 + sin 2 A ) cos 2 A − cos 2 A 2 sin 2 A = cos 2 A 1 + sin 2 A − cos 2 A 2 sin 2 A = cos 2 A 1 + sin 2 A − 2 sin 2 A = cos 2 A 1 − sin 2 A = cos 2 A cos 2 A = 1.
Since, L.H.S. = R.H.S. hence proved that sec4 A(1 - sin4 A) - 2tan2 A = 1.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
1 sin A + cos A + 1 + 1 sin A + cos A - 1 = sec A + cosec A . \dfrac{1}{\text{sin A + cos A + 1}} + \dfrac{1}{\text{sin A + cos A - 1}} = \text{sec A + cosec A}. sin A + cos A + 1 1 + sin A + cos A - 1 1 = sec A + cosec A .
Answer
Solving L.H.S.,
⇒ 1 sin A + cos A + 1 + 1 sin A + cos A - 1 = sin A + cos A - 1 + sin A + cos A + 1 (sin A + cos A + 1)(sin A + cos A - 1) = 2 (sin A + cos A) (sin A + cos A) 2 − 1 = 2 (sin A + cos A) sin 2 A + cos 2 A + 2 sin A cos A − 1 = 2 sin A + 2 cos A 1 − 1 + 2 sin A cos A = 2 sin A + 2 cos A 2 sin A cos A = 2 sin A 2 sin A cos A + 2 cos A 2 sin A cos A = 1 cos A + 1 sin A = sec A + cosec A . \Rightarrow \dfrac{1}{\text{sin A + cos A + 1}} + \dfrac{1}{\text{sin A + cos A - 1}} \\[1em] = \dfrac{\text{sin A + cos A - 1 + sin A + cos A + 1}}{\text{(sin A + cos A + 1)}\text{(sin A + cos A - 1)}} \\[1em] = \dfrac{2\text{(sin A + cos A)}}{\text{(sin A + cos A)}^2 - 1} \\[1em] = \dfrac{2\text{(sin A + cos A)}}{\text{sin}^2 A + \text{cos}^2 A + \text{2 sin A cos A} - 1} \\[1em] = \dfrac{\text{2 sin A + 2 cos A}}{1 - 1 + \text{2 sin A cos A}} \\[1em] = \dfrac{\text{2 sin A + 2 cos A}}{\text{2 sin A cos A}} \\[1em] = \dfrac{\text{2 sin A}}{\text{2 sin A cos A}} + \dfrac{\text{2 cos A}}{\text{2 sin A cos A}} \\[1em] = \dfrac{1}{\text{cos A}} + \dfrac{1}{\text{sin A}} \\[1em] = \text{sec A + cosec A}. ⇒ sin A + cos A + 1 1 + sin A + cos A - 1 1 = (sin A + cos A + 1) (sin A + cos A - 1) sin A + cos A - 1 + sin A + cos A + 1 = (sin A + cos A) 2 − 1 2 (sin A + cos A) = sin 2 A + cos 2 A + 2 sin A cos A − 1 2 (sin A + cos A) = 1 − 1 + 2 sin A cos A 2 sin A + 2 cos A = 2 sin A cos A 2 sin A + 2 cos A = 2 sin A cos A 2 sin A + 2 sin A cos A 2 cos A = cos A 1 + sin A 1 = sec A + cosec A .
Since, L.H.S. = R.H.S. hence proved that 1 sin A + cos A + 1 + 1 sin A + cos A - 1 = sec A + cosec A \dfrac{1}{\text{sin A + cos A + 1}} + \dfrac{1}{\text{sin A + cos A - 1}} = \text{sec A + cosec A} sin A + cos A + 1 1 + sin A + cos A - 1 1 = sec A + cosec A
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
sin 3 θ + cos 3 θ sin θ + cos θ + sin θ cos θ = 1. \dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1. sin θ + cos θ sin 3 θ + cos 3 θ + sin θ cos θ = 1.
Answer
Solving L.H.S.,
⇒ sin 3 θ + cos 3 θ sin θ + cos θ + sin θ cos θ = (sin θ + cos θ) ( sin 2 θ + cos 2 θ − sin θ cos θ ) sin θ + cos θ + sin θ cos θ = sin 2 θ + cos 2 θ − sin θ cos θ + sin θ cos θ = 1. \Rightarrow \dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \dfrac{\text{(sin θ + cos θ)}(\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ})}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ + sin θ cos θ} \\[1em] = 1. ⇒ sin θ + cos θ sin 3 θ + cos 3 θ + sin θ cos θ = sin θ + cos θ (sin θ + cos θ) ( sin 2 θ + cos 2 θ − sin θ cos θ ) + sin θ cos θ = sin 2 θ + cos 2 θ − sin θ cos θ + sin θ cos θ = 1.
Since, L.H.S. = R.H.S. hence proved that sin 3 θ + cos 3 θ sin θ + cos θ + sin θ cos θ = 1 \dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1 sin θ + cos θ sin 3 θ + cos 3 θ + sin θ cos θ = 1
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
(sec A - tan A)2 (1 + sin A) = 1 - sin A.
Answer
Solving L.H.S.,
⇒ ( 1 cos A − sin A cos A ) 2 ( 1 + sin A ) = ( 1 - sin A cos A ) 2 (1 + sin A) = (1 - sin A) 2 (1 + sin A) 1 - sin 2 A = (1 - sin A) 2 (1 + sin A) (1 - sin A)(1 + sin A) = 1 - sin A . \Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2(1 + \text{sin A}) \\[1em] = \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] = \dfrac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{1 - sin}^2 A} \\[1em] = \dfrac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \text{1 - sin A}. ⇒ ( cos A 1 − cos A sin A ) 2 ( 1 + sin A ) = ( cos A 1 - sin A ) 2 (1 + sin A) = 1 - sin 2 A (1 - sin A) 2 (1 + sin A) = (1 - sin A)(1 + sin A) (1 - sin A) 2 (1 + sin A) = 1 - sin A .
Since, L.H.S. = R.H.S. hence proved that (sec A - tan A)2 (1 + sin A) = 1 - sin A.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
cos A 1 - tan A − sin 2 A cos A - sin A = sin A + cos A . \dfrac{\text{cos A}}{\text{1 - tan A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A}. 1 - tan A cos A − cos A - sin A sin 2 A = sin A + cos A .
Answer
Solving L.H.S.,
⇒ cos A 1 − sin A cos A − sin 2 A cos A - sin A = cos A cos A - sin A cos A − sin 2 A cos A - sin A = cos 2 A cos A - sin A − sin 2 A cos A - sin A = cos 2 A − sin 2 A cos A - sin A = (cos A - sin A)(cos A + sin A) (cos A - sin A) = cos A + sin A . \Rightarrow \dfrac{\text{cos A}}{1 - \dfrac{\text{sin A}}{\text{cos A}}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos A}}{\dfrac{\text{cos A - sin A}}{\text{cos A}}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos}^2 A}{\text{cos A - sin A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{(cos A - sin A)(cos A + sin A)}}{\text{(cos A - sin A)}} \\[1em] = \text{cos A + sin A}. ⇒ 1 − cos A sin A cos A − cos A - sin A sin 2 A = cos A cos A - sin A cos A − cos A - sin A sin 2 A = cos A - sin A cos 2 A − cos A - sin A sin 2 A = cos A - sin A cos 2 A − sin 2 A = (cos A - sin A) (cos A - sin A)(cos A + sin A) = cos A + sin A .
Since, L.H.S. = R.H.S. hence proved that cos A 1 - tan A − sin 2 A cos A - sin A = sin A + cos A \dfrac{\text{cos A}}{\text{1 - tan A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A} 1 - tan A cos A − cos A - sin A sin 2 A = sin A + cos A
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
(sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.
Answer
Solving L.H.S.,
⇒ (sec A - cosec A)(1 + tan A + cot A) = ( 1 cos A − 1 sin A ) ( 1 + sin A cos A + cos A sin A ) = ( sin A - cos A sin A cos A ) ( sin A cos A + sin 2 A + cos 2 A sin A cos A ) = ( sin A - cos A ) ( sin A cos A + 1 ) sin 2 A cos 2 A \Rightarrow \text{(sec A - cosec A)(1 + tan A + cot A)} \\[1em] = \Big(\dfrac{1}{\text{cos A}} - \dfrac{1}{\text{sin A}}\Big)\Big(1 + \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big) \\[1em] = \Big(\dfrac{\text{sin A - cos A}}{\text{sin A cos A}}\Big)\Big(\dfrac{\text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \Big) \\[1em] = \dfrac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A} \\[1em] ⇒ (sec A - cosec A)(1 + tan A + cot A) = ( cos A 1 − sin A 1 ) ( 1 + cos A sin A + sin A cos A ) = ( sin A cos A sin A - cos A ) ( sin A cos A sin A cos A + sin 2 A + cos 2 A ) = sin 2 A cos 2 A ( sin A - cos A ) ( sin A cos A + 1 )
Now solving R.H.S.,
⇒ tan A sec A - cot A cosec A = sin A cos A × 1 cos A − cos A sin A × 1 sin A = sin A cos 2 A − cos A sin 2 A = sin 3 A − cos 3 A sin 2 A cos 2 A = (sin A - cos A)(sin 2 A + cos 2 A + sin A cos A ) sin 2 A cos 2 A = ( sin A - cos A ) ( sin A cos A + 1 ) sin 2 A cos 2 A . \Rightarrow \text{tan A sec A - cot A cosec A} \\[1em] = \dfrac{\text{sin A}}{\text{cos A}} \times \dfrac{1}{\text{cos A}} - \dfrac{\text{cos A}}{\text{sin A}} \times \dfrac{1}{\text{sin A}} \\[1em] = \dfrac{\text{sin A}}{\text{cos}^2 A} - \dfrac{\text{cos A}}{\text{sin}^2 A} \\[1em] = \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \dfrac{\text{(sin A - cos A)(sin}^2 A + \text{cos}^2 A + \text{sin A cos A})}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \dfrac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A}. ⇒ tan A sec A - cot A cosec A = cos A sin A × cos A 1 − sin A cos A × sin A 1 = cos 2 A sin A − sin 2 A cos A = sin 2 A cos 2 A sin 3 A − cos 3 A = sin 2 A cos 2 A (sin A - cos A)(sin 2 A + cos 2 A + sin A cos A ) = sin 2 A cos 2 A ( sin A - cos A ) ( sin A cos A + 1 ) .
Since, L.H.S. = R.H.S. hence proved that (sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
tan 2 θ tan 2 θ − 1 + cosec 2 θ sec 2 θ − cosec 2 θ = 1 sin 2 θ − cos 2 θ \dfrac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \dfrac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} tan 2 θ − 1 tan 2 θ + sec 2 θ − cosec 2 θ cosec 2 θ = sin 2 θ − cos 2 θ 1
Answer
Solving L.H.S.,
⇒ sin 2 θ cos 2 θ sin 2 θ cos 2 θ − 1 + 1 sin 2 θ 1 cos 2 θ − 1 sin 2 θ = sin 2 θ cos 2 θ sin 2 θ − cos 2 θ cos 2 θ + 1 sin 2 θ sin 2 θ − cos 2 θ cos 2 θ sin 2 θ = sin 2 θ sin 2 θ − cos 2 θ + 1 sin 2 θ − cos 2 θ cos 2 θ = sin 2 θ sin 2 θ − cos 2 θ + cos 2 θ sin 2 θ − cos 2 θ = sin 2 θ + cos 2 θ sin 2 θ − cos 2 θ = 1 sin 2 θ − cos 2 θ . \Rightarrow \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}} - 1} + \dfrac{\dfrac{1}{\text{sin}^2 \text{ θ}}}{\dfrac{1}{\text{cos}^2 \text{ θ}} - \dfrac{1}{\text{sin}^2 \text{ θ}}} \\[1em] = \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} + \dfrac{\dfrac{1}{\text{sin}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ} \text{ sin}^2 \text{ θ}}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \dfrac{1}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}. ⇒ cos 2 θ sin 2 θ − 1 cos 2 θ sin 2 θ + cos 2 θ 1 − sin 2 θ 1 sin 2 θ 1 = cos 2 θ sin 2 θ − cos 2 θ cos 2 θ sin 2 θ + cos 2 θ sin 2 θ sin 2 θ − cos 2 θ sin 2 θ 1 = sin 2 θ − cos 2 θ sin 2 θ + cos 2 θ sin 2 θ − cos 2 θ 1 = sin 2 θ − cos 2 θ sin 2 θ + sin 2 θ − cos 2 θ cos 2 θ = sin 2 θ − cos 2 θ sin 2 θ + cos 2 θ = sin 2 θ − cos 2 θ 1 .
Since, L.H.S. = R.H.S. hence proved that,
tan 2 θ tan 2 θ − 1 + cosec 2 θ sec 2 θ − cosec 2 θ = 1 sin 2 θ − cos 2 θ \dfrac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \dfrac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} tan 2 θ − 1 tan 2 θ + sec 2 θ − cosec 2 θ cosec 2 θ = sin 2 θ − cos 2 θ 1
sin A + cos A sin A - cos A + sin A - cos A sin A + cos A = 2 sin 2 A − cos 2 A = 2 1 − 2 cos 2 A = 2 sec 2 A tan 2 A − 1 \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} = \dfrac{2}{1 - 2\text{cos }^2A} = \dfrac{2\text{ sec }^2A}{\text{tan }^2A - 1} sin A - cos A sin A + cos A + sin A + cos A sin A - cos A = sin 2 A − cos 2 A 2 = 1 − 2 cos 2 A 2 = tan 2 A − 1 2 sec 2 A
Answer
Given,
⇒ sin A + cos A sin A - cos A + sin A - cos A sin A + cos A ⇒ (sin A + cos A) 2 + (sin A - cos A) 2 ( sin A − cos A ) ( sin A + cos A ) ⇒ sin 2 A + cos 2 A + 2 sin A.cos A + sin 2 A + cos 2 A − 2 sin A.cos A sin 2 A − cos 2 A ⇒ 2 sin 2 A + 2 cos 2 A sin 2 A − cos 2 A ⇒ 2 ( 1 − cos 2 A ) + 2 cos 2 A sin 2 A − cos 2 A ⇒ 2 − 2 cos 2 A + 2 cos 2 A sin 2 A − cos 2 A ⇒ 2 sin 2 A − cos 2 A \Rightarrow \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}}\\[1em] \Rightarrow \dfrac{\text{(sin A + cos A)}^2 + \text{(sin A - cos A)}^2}{(\text{sin} A - \text{cos} A)(\text{sin A + cos A})}\\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 2\text{sin A.cos A} + \text{sin}^2 A + \text{cos}^2 A - 2\text{sin A.cos A}}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2\text{sin}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2(1 - \text{cos}^2 A) + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2 - 2\text{cos}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A} ⇒ sin A - cos A sin A + cos A + sin A + cos A sin A - cos A ⇒ ( sin A − cos A ) ( sin A + cos A ) (sin A + cos A) 2 + (sin A - cos A) 2 ⇒ sin 2 A − cos 2 A sin 2 A + cos 2 A + 2 sin A.cos A + sin 2 A + cos 2 A − 2 sin A.cos A ⇒ sin 2 A − cos 2 A 2 sin 2 A + 2 cos 2 A ⇒ sin 2 A − cos 2 A 2 ( 1 − cos 2 A ) + 2 cos 2 A ⇒ sin 2 A − cos 2 A 2 − 2 cos 2 A + 2 cos 2 A ⇒ sin 2 A − cos 2 A 2
So proved,
sin A + cos A sin A - cos A + sin A - cos A sin A + cos A = 2 sin 2 A − cos 2 A \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} sin A - cos A sin A + cos A + sin A + cos A sin A - cos A = sin 2 A − cos 2 A 2
Now,
⇒ 2 sin 2 A − cos 2 A ⇒ 2 ( 1 − cos 2 A ) − cos 2 A ⇒ 2 1 − cos 2 A − cos 2 A ⇒ 2 1 − 2 cos 2 A \Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{(1 - \text{cos}^2 A) - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{1 - \text{cos}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{1 - 2\text{cos}^2 A} ⇒ sin 2 A − cos 2 A 2 ⇒ ( 1 − cos 2 A ) − cos 2 A 2 ⇒ 1 − cos 2 A − cos 2 A 2 ⇒ 1 − 2 cos 2 A 2
So proved,
sin A + cos A sin A - cos A + sin A - cos A sin A + cos A = 2 1 − 2 cos 2 A \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{1 - 2\text{cos}^2 A} sin A - cos A sin A + cos A + sin A + cos A sin A - cos A = 1 − 2 cos 2 A 2
Now,
⇒ 2 sin 2 A − cos 2 A \Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A} ⇒ sin 2 A − cos 2 A 2
Dividing numerator and denominator by cos2 A, we get :
⇒ 2 cos 2 A sin 2 A cos 2 A − cos 2 A cos 2 A ⇒ 2 sec 2 A tan 2 A − 1 \Rightarrow \dfrac{\dfrac{2}{\text{cos}^2 A}}{\dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \dfrac{\text{cos}^2 A}{\text{cos}^2 A}}\\[1em] \Rightarrow \dfrac{2\text{sec}^2 A}{\text{tan}^2 A - 1}\\[1em] ⇒ cos 2 A sin 2 A − cos 2 A cos 2 A cos 2 A 2 ⇒ tan 2 A − 1 2 sec 2 A
So proved,
sin A + cos A sin A - cos A + sin A - cos A sin A + cos A = 2 sec 2 A tan 2 A − 1 \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2\text{sec }^2A}{\text{tan }^2A - 1} sin A - cos A sin A + cos A + sin A + cos A sin A - cos A = tan 2 A − 1 2 sec 2 A
Hence, proved that
sin A + cos A sin A - cos A + sin A - cos A sin A + cos A = 2 sin 2 A − cos 2 A = 2 1 − 2 cos 2 A = 2 sec 2 A tan 2 A − 1 \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} = \dfrac{2}{1 - 2\text{cos }^2A} = \dfrac{2\text{sec }^2A}{\text{tan }^2A - 1} sin A - cos A sin A + cos A + sin A + cos A sin A - cos A = sin 2 A − cos 2 A 2 = 1 − 2 cos 2 A 2 = tan 2 A − 1 2 sec 2 A
Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
2(sin6 θ + cos6 θ) - 3(sin4 θ + cos4 θ) + 1 = 0.
Answer
Solving L.H.S.,
⇒ 2[(sin2 θ)3 + (cos2 θ)3 ] - 3[(sin2 θ)2 + (cos2 θ)2 ] + 1
⇒ 2[(sin2 θ + cos2 θ)3 - 3sin2 θ cos2 θ(sin2 θ + cos2 θ)] - 3[(sin2 θ + cos2 θ)2 - 2sin2 θ cos2 θ)] + 1
⇒ 2[(1)3 - 3sin2 θ cos2 θ(1)] - 3[(1)2 - 2sin2 θ cos2 θ)] + 1
⇒ 2 - 6sin2 θ cos2 θ - 3 + 6sin2 θ cos2 θ + 1
⇒ 2 - 3 + 1 - 6sin2 θ cos2 θ + 6sin2 θ cos2 θ
⇒ 3 - 3
⇒ 0.
Since, L.H.S. = R.H.S. hence, proved that 2(sin6 θ + cos6 θ) - 3(sin4 θ + cos4 θ) + 1 = 0.
If cot θ + cos θ = m and cot θ - cos θ = n, then prove that (m2 - n2 )2 = 16 mn.
Answer
Given,
cot θ + cos θ = m ....(i)
Adding (i) and (ii) we get,
⇒ m + n = cot θ + cos θ + cot θ - cos θm + n 2 \dfrac{m + n}{2} 2 m + n
∴ tan θ = 2 m + n \dfrac{2}{m + n} m + n 2
Subtracting (ii) from (i) we get,
m - n = cot θ + cos θ - cot θ + cos θm − n 2 \dfrac{m - n}{2} 2 m − n
∴ sec θ = 2 m − n \dfrac{2}{m - n} m − n 2
Squaring and subtracting (iii) from (iv),
⇒ sec 2 θ − tan 2 θ = ( 2 m − n ) 2 − ( 2 m + n ) 2 ⇒ 1 = 4 ( m − n ) 2 − 4 ( m + n ) 2 ⇒ 4 [ 1 ( m − n ) 2 − 1 ( m + n ) 2 ] = 1 ⇒ 4 [ ( m + n ) 2 − ( m − n ) 2 ( m + n ) 2 ( m − n ) 2 ] = 1 ⇒ 4 [ m 2 + n 2 + 2 m n − m 2 − n 2 + 2 m n ( m + n ) ( m − n ) ( m + n ) ( m − n ) ] = 1 ⇒ 4 [ m 2 + n 2 + 2 m n − m 2 − n 2 + 2 m n ( m 2 − n 2 ) ( m 2 − n 2 ) ] = 1 ⇒ 4 × 4 m n ( m 2 − n 2 ) 2 = 1 ⇒ 16 m n = ( m 2 − n 2 ) 2 . \Rightarrow \text{sec}^2 \text{ θ} - \text{tan}^2 \text{ θ} = \Big(\dfrac{2}{m - n}\Big)^2 - \Big(\dfrac{2}{m + n}\Big)^2 \\[1em] \Rightarrow 1 = \dfrac{4}{(m - n)^2} - \dfrac{4}{(m + n)^2} \\[1em] \Rightarrow 4\Big[\dfrac{1}{(m - n)^2} - \dfrac{1}{(m + n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{(m + n)^2 - (m - n)^2}{(m + n)^2(m - n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{m^2 + n^2 + 2mn - m^2 - n^2 + 2mn}{(m + n)(m - n)(m + n)(m - n)} \Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{\cancel{m^2} + \cancel{n^2} + 2mn - \cancel{m^2} - \cancel{n^2} + 2mn}{(m^2 - n^2)(m^2 - n^2)} \Big] = 1 \\[1em] \Rightarrow 4 \times \dfrac{4mn}{(m^2 - n^2)^2} = 1 \\[1em] \Rightarrow 16 mn = (m^2 - n^2)^2. ⇒ sec 2 θ − tan 2 θ = ( m − n 2 ) 2 − ( m + n 2 ) 2 ⇒ 1 = ( m − n ) 2 4 − ( m + n ) 2 4 ⇒ 4 [ ( m − n ) 2 1 − ( m + n ) 2 1 ] = 1 ⇒ 4 [ ( m + n ) 2 ( m − n ) 2 ( m + n ) 2 − ( m − n ) 2 ] = 1 ⇒ 4 [ ( m + n ) ( m − n ) ( m + n ) ( m − n ) m 2 + n 2 + 2 mn − m 2 − n 2 + 2 mn ] = 1 ⇒ 4 [ ( m 2 − n 2 ) ( m 2 − n 2 ) m 2 + n 2 + 2 mn − m 2 − n 2 + 2 mn ] = 1 ⇒ 4 × ( m 2 − n 2 ) 2 4 mn = 1 ⇒ 16 mn = ( m 2 − n 2 ) 2 .
Hence, proved that (m2 - n2 )2 = 16 mn.
When 0° < θ < 90°, solve the following equation:
2 cos2 θ + sin θ - 2 = 0
Answer
Given,
2 cos2 θ + sin θ - 2 = 0
On Solving,
⇒ 2(1 - sin2 θ) + sin θ - 2 = 0
= 2 - 2 sin2 θ + sin θ - 2 = 0
= sin θ -2 sin2 θ = 0
= sin θ (1 - 2 sin θ) = 0
So, either sin θ = 0 or 1 - 2 sin θ = 0
If, sin θ = 0
Given, θ > 0° hence, θ = 0° is not possible.
∴ 1 - 2 sin θ = 0
⇒ 1 = 2 sin θ
⇒ sin θ = 1 2 \dfrac{1}{2} 2 1
⇒ sin θ = 30°.
Hence, the value of θ = 30°.
When 0° < θ < 90°, solve the following equation:
3 cos θ = 2 sin2 θ
Answer
Given,
3 cos θ = 2 sin2 θ
On Solving,
⇒ 3 cos θ = 2(1 - cos2 θ)
⇒ 3 cos θ = 2 - 2cos2 θ
⇒ 2 cos2 θ + 3 cos θ - 2 = 0
⇒ 2 cos2 θ + 4 cos θ - cos θ - 2 = 0
⇒ 2 cos θ(cos θ + 2) - 1(cos θ + 2) = 0
⇒ (2 cos θ - 1)(cos θ + 2) = 0
⇒ 2 cos θ - 1 = 0 or cos θ + 2 = 0
⇒ cos θ = 1 2 \dfrac{1}{2} 2 1
But cos θ = -2 is not possible.
∴ cos θ = 1 2 \dfrac{1}{2} 2 1
⇒ cos θ = cos 60°
⇒ θ = 60°.
Hence, the value of θ = 60°.
When 0° < θ < 90°, solve the following equation:
sec2 θ - 2 tan θ = 0
Answer
Given,
sec2 θ - 2 tan θ = 0
On Solving,
⇒ 1 + tan2 θ - 2 tan θ = 0
⇒ tan2 θ - 2 tan θ + 1 = 0
⇒ (tan θ - 1)2 = 0
⇒ tan θ - 1 = 0
⇒ tan θ = 1
⇒ tan θ = tan 45°
⇒ θ = 45°.
Hence, the value of θ = 45°.
When 0° < θ < 90°, solve the following equation:
tan2 θ = 3 (sec θ - 1).
Answer
Given,
tan2 θ = 3 (sec θ - 1)
On Solving,
⇒ sec2 θ - 1 = 3 sec θ - 3
⇒ sec2 θ - 1 - 3 sec θ + 3 = 0
⇒ sec2 θ - 3 sec θ + 2 = 0
⇒ sec2 θ - 2 sec θ - sec θ + 2 = 0
⇒ sec θ (sec θ - 2) - 1(sec θ - 2) = 0
⇒ (sec θ - 1)(sec θ - 2) = 0
⇒ sec θ - 1 = 0 or sec θ - 2 = 0
⇒ sec θ = 1 or sec θ = 2.
If, sec θ = 1
Given, θ > 0° hence, θ = 0° is not possible.
∴ sec θ = 2
⇒ sec θ = sec 60°
⇒ θ = 60°.
Hence, the value of θ = 60°.
