Without using trigonometric tables, evaluate the following:
cos 18° sin 72° \dfrac{\text{cos 18°}}{\text{sin 72°}} sin 72° cos 18°
Answer
(i) Solving,
⇒ cos 18° sin 72° ⇒ cos 18° sin (90° - 18°) As, sin (90 - θ) = cos θ ⇒ cos 18° cos 18° ⇒ 1. \Rightarrow \dfrac{\text{cos 18°}}{\text{sin 72°}} \\[1em] \Rightarrow \dfrac{\text{cos 18°}}{\text{sin (90° - 18°)}} \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cos 18°}}{\text{cos 18°}} \\[1em] \Rightarrow 1. ⇒ sin 72° cos 18° ⇒ sin (90° - 18°) cos 18° As, sin (90 - θ) = cos θ ⇒ cos 18° cos 18° ⇒ 1.
Hence, cos 18° sin 72° \dfrac{\text{cos 18°}}{\text{sin 72°}} sin 72° cos 18°
Without using trigonometric tables, evaluate the following:
tan 41° cot 49° \dfrac{\text{tan 41°}}{\text{cot 49°}} cot 49° tan 41°
Answer
(ii) Solving,
⇒ tan 41° cot 49° ⇒ tan (90° - 49°) cot 49° As, tan (90 - θ) = cot θ ⇒ cot 49° cot 49° ⇒ 1. \Rightarrow \dfrac{\text{tan 41°}}{\text{cot 49°}} \\[1em] \Rightarrow \dfrac{\text{tan (90° - 49°)}}{\text{cot 49°}} \\[1em] \text{As, tan (90 - θ) = cot θ} \\[1em] \Rightarrow \dfrac{\text{cot 49°}}{\text{cot 49°}} \\[1em] \Rightarrow 1. ⇒ cot 49° tan 41° ⇒ cot 49° tan (90° - 49°) As, tan (90 - θ) = cot θ ⇒ cot 49° cot 49° ⇒ 1.
Hence, tan 41° cot 49° = 1 \dfrac{\text{tan 41°}}{\text{cot 49°}} = 1 cot 49° tan 41° = 1
Without using trigonometric tables, evaluate the following:
cosec 17°30 ′ sec 72° 30 ′ \dfrac{\text{cosec 17°30}'}{\text{sec 72° 30}'} sec 72° 30 ′ cosec 17°30 ′
Answer
Solving,
⇒ cosec 17° 30 ′ sec 72° 30 ′ ⇒ cosec (90° - 72° 30 ′ ) sec 72° 30 ′ As, cosec (90 - θ) = sec θ ⇒ sec 72° 30 ′ sec 72° 30 ′ ⇒ 1. \Rightarrow \dfrac{\text{cosec 17° 30}'}{\text{sec 72° 30}'} \\[1em] \Rightarrow \dfrac{\text{cosec (90° - 72° 30}')}{\text{sec 72° 30}'} \\[1em] \text{As, cosec (90 - θ) = sec θ} \\[1em] \Rightarrow \dfrac{\text{sec 72° 30}'}{\text{sec 72° 30}'} \\[1em] \Rightarrow 1. ⇒ sec 72° 30 ′ cosec 17° 30 ′ ⇒ sec 72° 30 ′ cosec (90° - 72° 30 ′ ) As, cosec (90 - θ) = sec θ ⇒ sec 72° 30 ′ sec 72° 30 ′ ⇒ 1.
Hence, cosec 17° 30 ′ sec 72° 30 ′ = 1 \dfrac{\text{cosec 17° 30}'}{\text{sec 72° 30}'} = 1 sec 72° 30 ′ cosec 17° 30 ′ = 1
Without using trigonometric tables, evaluate the following:
cot 40° tan 50° − 1 2 ( cos 35° sin 55° ) \dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big) tan 50° cot 40° − 2 1 ( sin 55° cos 35° )
Answer
Solving,
⇒ cot 40° tan 50° − 1 2 ( cos 35° sin 55° ) ⇒ cot 40° tan (90° - 40°) − 1 2 ( cos 35° sin (90° - 35°) ) As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ ⇒ cot 40° cot 40° − 1 2 ( cos 35° cos 35° ) ⇒ 1 − 1 2 ⇒ 1 2 . \Rightarrow \dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big) \\[1em] \Rightarrow \dfrac{\text{cot 40°}}{\text{tan (90° - 40°)}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin (90° - 35°)}}\Big) \\[1em] \text{As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cot 40°}}{\text{cot 40°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{cos 35°}}\Big) \\[1em] \Rightarrow 1 - \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{2}. ⇒ tan 50° cot 40° − 2 1 ( sin 55° cos 35° ) ⇒ tan (90° - 40°) cot 40° − 2 1 ( sin (90° - 35°) cos 35° ) As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ ⇒ cot 40° cot 40° − 2 1 ( cos 35° cos 35° ) ⇒ 1 − 2 1 ⇒ 2 1 .
Hence, cot 40° tan 50° − 1 2 ( cos 35° sin 55° ) = 1 2 \dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big) = \dfrac{1}{2} tan 50° cot 40° − 2 1 ( sin 55° cos 35° ) = 2 1
Without using trigonometric tables, evaluate the following:
( sin 49° cos 41° ) 2 + ( cos 41° sin 49° ) 2 \Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 ( cos 41° sin 49° ) 2 + ( sin 49° cos 41° ) 2
Answer
Solving,
⇒ ( sin 49° cos 41° ) 2 + ( cos 41° sin 49° ) 2 ⇒ ( sin (90° - 41°) cos 41° ) 2 + ( cos 41° sin (90° - 41°) ) 2 As, sin (90 - θ) = cos θ ⇒ ( cos 41° cos 41° ) 2 + ( cos 41° cos 41° ) 2 ⇒ 1 2 + 1 2 ⇒ 1 + 1 ⇒ 2. \Rightarrow \Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin (90° - 41°)}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin (90° - 41°)}}\Big)^2 \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \Big(\dfrac{\text{cos 41°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{cos 41°}}\Big)^2 \\[1em] \Rightarrow 1^2 + 1^2 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ ( cos 41° sin 49° ) 2 + ( sin 49° cos 41° ) 2 ⇒ ( cos 41° sin (90° - 41°) ) 2 + ( sin (90° - 41°) cos 41° ) 2 As, sin (90 - θ) = cos θ ⇒ ( cos 41° cos 41° ) 2 + ( cos 41° cos 41° ) 2 ⇒ 1 2 + 1 2 ⇒ 1 + 1 ⇒ 2.
Hence, ( sin 49° cos 41° ) 2 + ( cos 41° sin 49° ) 2 = 2. \Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 = 2. ( cos 41° sin 49° ) 2 + ( sin 49° cos 41° ) 2 = 2.
Without using trigonometric tables, evaluate the following:
sin 72° cos 18° − sec 32° cosec 58° \dfrac{\text{sin 72°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} cos 18° sin 72° − cosec 58° sec 32°
Answer
Solving,
⇒ sin 72° cos 18° − sec 32° cosec 58° ⇒ sin (90° - 18°) cos 18° − sec 32° cosec (90° - 32°) As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ ⇒ cos 18° cos 18° − sec 32° sec 32° ⇒ 1 − 1 ⇒ 0. \Rightarrow \dfrac{\text{sin 72°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} \\[1em] \Rightarrow \dfrac{\text{sin (90° - 18°)}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec (90° - 32°)}} \\[1em] \text{As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cos 18°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{sec 32°}} \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0. ⇒ cos 18° sin 72° − cosec 58° sec 32° ⇒ cos 18° sin (90° - 18°) − cosec (90° - 32°) sec 32° As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ ⇒ cos 18° cos 18° − sec 32° sec 32° ⇒ 1 − 1 ⇒ 0.
Hence, sin 72° cos 18° − sec 32° cosec 58° = 0. \dfrac{\text{sin 72°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} = 0. cos 18° sin 72° − cosec 58° sec 32° = 0.
Without using trigonometric tables, evaluate the following:
cos 75° sin 15° + sin 12° cos 78° − cos 18° sin 72° \dfrac{\text{cos 75°}}{\text{sin 15°}} + \dfrac{\text{sin 12°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin 72°}} sin 15° cos 75° + cos 78° sin 12° − sin 72° cos 18°
Answer
Solving,
⇒ cos 75° sin 15° + sin 12° cos 78° − cos 18° sin 72° ⇒ cos 75° sin (90° - 75°) + sin (90° - 78°) cos 78° − cos 18° sin (90° - 18°) As, sin (90 - θ) = cos θ ⇒ cos 75° cos 75° + cos 78° cos 78° − cos 18° cos 18° ⇒ 1 + 1 − 1 ⇒ 1. \Rightarrow \dfrac{\text{cos 75°}}{\text{sin 15°}} + \dfrac{\text{sin 12°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin 72°}} \\[1em] \Rightarrow \dfrac{\text{cos 75°}}{\text{sin (90° - 75°)}} + \dfrac{\text{sin (90° - 78°)}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin (90° - 18°)}} \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cos 75°}}{\text{cos 75°}} + \dfrac{\text{cos 78°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{cos 18°}} \\[1em] \Rightarrow 1 + 1 - 1 \\[1em] \Rightarrow 1. ⇒ sin 15° cos 75° + cos 78° sin 12° − sin 72° cos 18° ⇒ sin (90° - 75°) cos 75° + cos 78° sin (90° - 78°) − sin (90° - 18°) cos 18° As, sin (90 - θ) = cos θ ⇒ cos 75° cos 75° + cos 78° cos 78° − cos 18° cos 18° ⇒ 1 + 1 − 1 ⇒ 1.
Hence, cos 75° sin 15° + sin 12° cos 78° − cos 18° sin 72° = 1. \dfrac{\text{cos 75°}}{\text{sin 15°}} + \dfrac{\text{sin 12°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin 72°}} = 1. sin 15° cos 75° + cos 78° sin 12° − sin 72° cos 18° = 1.
Without using trigonometric tables, evaluate the following:
sin 25° sec 65° + cos 25° cosec 65° \dfrac{\text{sin 25°}}{\text{sec 65°}} + \dfrac{\text{cos 25°}}{\text{cosec 65°}} sec 65° sin 25° + cosec 65° cos 25°
Answer
Solving,
⇒ sin 25° sec 65° + cos 25° cosec 65° ⇒ sin 25° 1 cos 65° + cos 25° 1 sin 65° ⇒ sin 25°.cos 65° + cos 25°.sin 65° ⇒ sin 25°. cos(90° - 25°) + cos 25°. sin(90° - 25°) As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ ⇒ sin 25°. sin 25° + cos 25°. cos 25° ⇒ sin 2 25 ° + cos 2 25 ° ⇒ 1. \Rightarrow \dfrac{\text{sin 25°}}{\text{sec 65°}} + \dfrac{\text{cos 25°}}{\text{cosec 65°}} \\[1em] \Rightarrow \dfrac{\text{sin 25°}}{\dfrac{1}{\text{cos 65°}}} + \dfrac{\text{cos 25°}}{\dfrac{1}{\text{sin 65°}}} \\[1em] \Rightarrow \text{sin 25°.cos 65° + cos 25°.sin 65°} \\[1em] \Rightarrow \text{sin 25°. cos(90° - 25°) + cos 25°. sin(90° - 25°)} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \text{sin 25°. sin 25° + cos 25°. cos 25°} \\[1em] \Rightarrow \text{sin}^2 25° + \text{cos}^2 25° \\[1em] \Rightarrow 1. ⇒ sec 65° sin 25° + cosec 65° cos 25° ⇒ cos 65° 1 sin 25° + sin 65° 1 cos 25° ⇒ sin 25°.cos 65° + cos 25°.sin 65° ⇒ sin 25°. cos(90° - 25°) + cos 25°. sin(90° - 25°) As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ ⇒ sin 25°. sin 25° + cos 25°. cos 25° ⇒ sin 2 25° + cos 2 25° ⇒ 1.
Hence, sin 25° sec 65° + cos 25° cosec 65° = 1 \dfrac{\text{sin 25°}}{\text{sec 65°}} + \dfrac{\text{cos 25°}}{\text{cosec 65°}} = 1 sec 65° sin 25° + cosec 65° cos 25° = 1
Without using trigonometric tables, evaluate the following:
sin 62° - cos 28°
Answer
Solving,
⇒ sin 62° - cos 28°
⇒ sin (90° - 28°) - cos 28°
⇒ cos 28° - cos 28° [As, sin (90 - θ) = cos θ]
⇒ 0.
Hence, sin 62° - cos 28° = 0.
Without using trigonometric tables, evaluate the following:
cosec 35° - sec 55°
Answer
Solving,
⇒ cosec 35° - sec 55°
⇒ cosec (90° - 55°) - sec 55°
⇒ sec 55° - sec 55° [As, cosec (90 - θ) = sec θ]
⇒ 0.
Hence, cosec 35° - sec 55° = 0.
Without using trigonometric tables, evaluate the following:
cos2 26° + cos 64° sin 26° + tan 36° cot 54° \dfrac{\text{tan 36°}}{\text{cot 54°}} cot 54° tan 36°
Answer
Solving,
⇒ cos 2 26 ° + cos 64° sin 26° + tan 36° cot 54° ⇒ cos 2 26 ° + cos (90° - 26°) sin 26° + tan (90° - 54°) cot 54° As, cos (90 - θ) = sin θ and tan (90 - θ) = cot θ ⇒ cos 2 26 ° + sin 2 26 ° + cot 54° cot 54° As, sin 2 θ + cos 2 θ = 1 ⇒ 1 + 1 ⇒ 2. \Rightarrow \text{cos}^2 26° + \text{cos 64° sin 26°} + \dfrac{\text{tan 36°}}{\text{cot 54°}} \\[1em] \Rightarrow \text{cos}^2 26° + \text{cos (90° - 26°) sin 26°} + \dfrac{\text{tan (90° - 54°)}}{\text{cot 54°}} \\[1em] \text{As, cos (90 - θ) = sin θ and tan (90 - θ) = cot θ} \\[1em] \Rightarrow \text{cos}^2 26° + \text{sin}^2 26° + \dfrac{\text{cot 54°}}{\text{cot 54°}} \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ cos 2 26° + cos 64° sin 26° + cot 54° tan 36° ⇒ cos 2 26° + cos (90° - 26°) sin 26° + cot 54° tan (90° - 54°) As, cos (90 - θ) = sin θ and tan (90 - θ) = cot θ ⇒ cos 2 26° + sin 2 26° + cot 54° cot 54° As, sin 2 θ + cos 2 θ = 1 ⇒ 1 + 1 ⇒ 2.
Hence, cos2 26° + cos 64° sin 26° + tan 36° cot 54° \dfrac{\text{tan 36°}}{\text{cot 54°}} cot 54° tan 36°
Without using trigonometric tables, evaluate the following:
sec 17° cosec 73° + tan 68° cot 22° \dfrac{\text{sec 17°}}{\text{\text{cosec 73°}}} + \dfrac{\text{tan 68°}}{\text{cot 22°}} cosec 73° sec 17° + cot 22° tan 68° 2 44° + cos2 46°.
Answer
Solving,
⇒ sec 17° cosec (90° - 17°) + tan (90° - 22°) cot 22° + cos 2 44 ° + cos 2 ( 90 ° − 44 ° ) ⇒ sec 17° sec 17° + cot 22° cot 22° + cos 2 44 ° + sin 2 44 ° ⇒ 1 + 1 + 1 ⇒ 3. \Rightarrow \dfrac{\text{sec 17°}}{\text{cosec (90° - 17°)}} + \dfrac{\text{tan (90° - 22°)}}{\text{cot 22°}} + \text{cos}^2 44° + \text{cos}^2 (90° - 44°) \\[1em] \Rightarrow \dfrac{\text{sec 17°}}{\text{sec 17°}} + \dfrac{\text{cot 22°}}{\text{cot 22°}} + \text{cos}^2 44° + \text{sin}^2 44° \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3. ⇒ cosec (90° - 17°) sec 17° + cot 22° tan (90° - 22°) + cos 2 44° + cos 2 ( 90° − 44° ) ⇒ sec 17° sec 17° + cot 22° cot 22° + cos 2 44° + sin 2 44° ⇒ 1 + 1 + 1 ⇒ 3.
Hence, sec 17° cosec 73° + tan 68° cot 22° \dfrac{\text{sec 17°}}{\text{\text{cosec 73°}}} + \dfrac{\text{tan 68°}}{\text{cot 22°}} cosec 73° sec 17° + cot 22° tan 68° 2 44° + cos2 46° = 3.
Without using trigonometric tables, evaluate the following:
cos 65° sin 25° + cos 32° sin 58° − sin 28° sec 62° + cosec 2 30 ° \dfrac{\text{cos 65°}}{\text{sin 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62° + cosec}^2 30° sin 25° cos 65° + sin 58° cos 32° − sin 28° sec 62° + cosec 2 30°
Answer
Solving,
⇒ cos 65° sin (90° - 65°) + cos 32° sin (90° - 32°) − sin 28° × 1 cos 62° + 1 sin 2 30 ° ⇒ cos 65° cos 65° + cos 32° cos 32° − sin 28° × 1 cos 62° + 1 sin 2 30 ° ⇒ 1 + 1 − sin 28° × 1 c o s ( 90 ° − 28 ° ) + 1 ( 1 2 ) 2 ⇒ 2 − sin 28° sin 28° + 4 ⇒ 2 − 1 + 4 ⇒ 5. \Rightarrow \dfrac{\text{cos 65°}}{\text{sin (90° - 65°)}} + \dfrac{\text{cos 32°}}{\text{sin (90° - 32°)}} - \text{sin 28°} \times \dfrac{1}{\text{cos 62°}} + \dfrac{1}{\text{sin}^2 30°} \\[1em] \Rightarrow \dfrac{\text{cos 65°}}{\text{cos 65°}} + \dfrac{\text{cos 32°}}{\text{cos 32°}} - \text{sin 28°} \times \dfrac{1}{\text{cos 62°}} + \dfrac{1}{\text{sin}^2 30°} \\[1em] \Rightarrow 1 + 1 - \text{sin 28°} \times \dfrac{1}{cos(90° - 28°)} + \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} \\[1em] \Rightarrow 2 - \dfrac{\text{sin 28°}}{\text{sin 28°}} + 4 \\[1em] \Rightarrow 2 - 1 + 4 \\[1em] \Rightarrow 5. ⇒ sin (90° - 65°) cos 65° + sin (90° - 32°) cos 32° − sin 28° × cos 62° 1 + sin 2 30° 1 ⇒ cos 65° cos 65° + cos 32° cos 32° − sin 28° × cos 62° 1 + sin 2 30° 1 ⇒ 1 + 1 − sin 28° × cos ( 90° − 28° ) 1 + ( 2 1 ) 2 1 ⇒ 2 − sin 28° sin 28° + 4 ⇒ 2 − 1 + 4 ⇒ 5.
Hence, cos 65° sin 25° + cos 32° sin 58° − sin 28° sec 62° + cosec 2 30 ° \dfrac{\text{cos 65°}}{\text{sin 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62° + cosec}^2 30° sin 25° cos 65° + sin 58° cos 32° − sin 28° sec 62° + cosec 2 30°
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
Solving,
⇒ sec 29° cosec 61° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° − 3 (sin 2 38 ° + sin 2 52 ° ) ⇒ sec 29° cosec (90° - 29°) + 2 cot (90° - 82°) cot (90° - 73°) cot 45° cot 73° cot 82° − 3 (sin 2 38 ° + sin 2 ( 90 ° − 38 ° ) ) As, cosec(90° - θ) = sec θ, cot(90° - θ) = tan θ ⇒ sec 29° sec 29° + 2 tan 82° tan 73° cot 45° × 1 tan 73° × 1 tan 82° − 3 (sin 2 38 ° + cos 2 38 ° ) As, sin 2 θ + cos 2 θ = 1 ⇒ 1 + 2 cot 45° − 3 ⇒ 1 + 2 − 3 ⇒ 0. \Rightarrow \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°) \\[1em] \Rightarrow \dfrac{\text{sec 29°}}{\text{cosec (90° - 29°)}} + \text{2 cot (90° - 82°) cot (90° - 73°) cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 (90° - 38°)) \\[1em] \text{As, cosec(90° - θ) = sec θ, cot(90° - θ) = tan θ} \\[1em] \Rightarrow \dfrac{\text{sec 29°}}{\text{sec 29°}} + \text{2 tan 82° tan 73° cot 45°} \times \dfrac{1}{\text{tan 73°}} \times \dfrac{1}{\text{tan 82°}} - \text{3 (sin}^2 38° + \text{cos}^2 38°) \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \\[1em] \Rightarrow 1 + \text{2 cot 45°} - 3 \\[1em] \Rightarrow 1 + 2 - 3 \\[1em] \Rightarrow 0. ⇒ cosec 61° sec 29° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° − 3 (sin 2 38° + sin 2 52° ) ⇒ cosec (90° - 29°) sec 29° + 2 cot (90° - 82°) cot (90° - 73°) cot 45° cot 73° cot 82° − 3 (sin 2 38° + sin 2 ( 90° − 38° )) As, cosec(90° - θ) = sec θ, cot(90° - θ) = tan θ ⇒ sec 29° sec 29° + 2 tan 82° tan 73° cot 45° × tan 73° 1 × tan 82° 1 − 3 (sin 2 38° + cos 2 38° ) As, sin 2 θ + cos 2 θ = 1 ⇒ 1 + 2 cot 45° − 3 ⇒ 1 + 2 − 3 ⇒ 0.
Hence, 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° )
Express the following in terms of trigonometric ratios of angles between 0° to 45° :
tan 81° + cos 72°
Answer
Solving,
⇒ tan 81° + cos 72°
⇒ tan (90° - 9°) + cos (90° - 18°)
As, tan(90° - θ) = cot θ, cos(90° - θ) = sin θ
⇒ cot 9° + sin 18°.
Hence, tan 81° + cos 72° = cot 9° + sin 18°.
Express the following in terms of trigonometric ratios of angles between 0° to 45° :
cot 49° + cosec 87°.
Answer
Solving,
⇒ cot 49° + cosec 87°
⇒ cot (90° - 41°) + cosec (90° - 3°)
As, cot(90° - θ) = tan θ, cosec(90° - θ) = sec θ
⇒ tan 41° + sec 3°.
Hence, cot 49° + cosec 87° = tan 41° + sec 3°.
Without using trigonometric tables, prove that:
sin2 28° - cos2 62° = 0
Answer
To prove,
sin2 28° - cos2 62° = 0.
Solving, L.H.S. of the equation.
sin2 28° - cos2 62°
= sin2 28° - cos2 (90° - 28°)
= sin2 28° - sin2 28°
= 0.
Since, L.H.S. = R.H.S.
Hence, proved that sin2 28° - cos2 62° = 0.
Without using trigonometric tables, prove that:
cos2 25° + cos2 65° = 1
Answer
To prove,
cos2 25° + cos2 65° = 1.
Solving, L.H.S. of the equation.
cos2 25° + cos2 65°
= cos2 25° + cos2 (90° - 25°)
As, cos (90° - θ) = sin θ
= cos2 25° + sin2 25°
= 1 [∵ cos2 θ + sin2 θ = 1]
Since, L.H.S. = R.H.S.
Hence, proved that cos2 25° + cos2 65° = 1.
Without using trigonometric tables, prove that:
cosec2 67° - tan2 23° = 1
Answer
To prove,
cosec2 67° - tan2 23° = 1
Solving L.H.S. of the equation,
⇒ cosec 2 67 ° − tan 2 23 ° ⇒ 1 sin 2 67 ° − tan 2 ( 90 ° − 67 ° ) ⇒ 1 sin 2 67 ° − cot 2 67 ° ⇒ 1 sin 2 67 ° − cos 2 67 ° sin 2 67 ° ⇒ 1 - cos 2 67 ° sin 2 67 ° ⇒ sin 2 67 ° sin 2 67 ° [ ∵ 1 - cos 2 θ = sin 2 θ ] ⇒ 1. \phantom{\Rightarrow} \text{cosec}^2 67° - \text{tan}^2 23° \\[1em] \Rightarrow \dfrac{1}{\text{sin}^2 67°} - \text{tan}^2 (90° - 67°) \\[1em] \Rightarrow \dfrac{1}{\text{sin}^2 67°} - \text{cot}^2 67° \\[1em] \Rightarrow \dfrac{1}{\text{sin}^2 67°} - \dfrac{\text{cos}^2 67°}{\text{sin}^2 67°} \\[1em] \Rightarrow \dfrac{\text{1 - cos}^2 67°}{\text{sin}^2 67°} \\[1em] \Rightarrow \dfrac{\text{sin}^2 67°}{\text{sin}^2 67°} \space [\because \text{ 1 - cos}^2 \text{ θ} = \text{sin}^2 \text{ θ}] \\[1em] \Rightarrow 1. ⇒ cosec 2 67° − tan 2 23° ⇒ sin 2 67° 1 − tan 2 ( 90° − 67° ) ⇒ sin 2 67° 1 − cot 2 67° ⇒ sin 2 67° 1 − sin 2 67° cos 2 67° ⇒ sin 2 67° 1 - cos 2 67° ⇒ sin 2 67° sin 2 67° [ ∵ 1 - cos 2 θ = sin 2 θ ] ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that cosec2 67° - tan2 23° = 1.
Without using trigonometric tables, prove that:
sec2 22° - cot2 68° = 1
Answer
To prove,
sec2 22° - cot2 68° = 1
Solving, L.H.S. of the equation we get :
sec2 22° - cot2 68°
⇒ sec2 22° - cot2 (90° - 22°)
We know that,
cot (90 - θ) = tan θ
⇒ sec2 22° - tan2 22°
⇒ 1 cos 2 22 ° − sin 2 22 ° cos 2 22 ° ⇒ 1 - sin 2 22 ° cos 2 22 ° As, 1 - sin 2 θ = cos 2 θ ⇒ cos 2 22 ° cos 2 22 ° ⇒ 1. \Rightarrow \dfrac{1}{\text{cos}^2 22°} - \dfrac{\text{sin}^2 22°}{\text{cos}^2 22°} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 22°}{\text{cos}^2 22°} \\[1em] \text{As, 1 - sin}^2 θ = \text{cos}^2 θ \\[1em] \Rightarrow \dfrac{\text{cos}^2 22°}{\text{cos}^2 22°} \\[1em] \Rightarrow 1. ⇒ cos 2 22° 1 − cos 2 22° sin 2 22° ⇒ cos 2 22° 1 - sin 2 22° As, 1 - sin 2 θ = cos 2 θ ⇒ cos 2 22° cos 2 22° ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that sec2 22° - cot2 68° = 1.
Without using trigonometric tables, prove that:
sin 63° cos 27° + cos 63° sin 27° = 1
Answer
To prove,
sin 63° cos 27° + cos 63° sin 27° = 1
Solving, L.H.S. of the equation we get,
⇒ sin (90° - 27°) cos 27° + cos (90° - 27°) sin 27°
We know that,
sin (90 - θ) = cos θ and cos (90 - θ) = sin θ
⇒ cos 27° cos 27° + sin 27° sin 27°
⇒ cos2 27° + sin2 27°
⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that sin 63° cos 27° + cos 63° sin 27° = 1.
Without using trigonometric tables, prove that:
sec 31° sin 59° + cos 31° cosec 59° = 2.
Answer
To prove,
sec 31° sin 59° + cos 31° cosec 59° = 2
Solving, L.H.S. of the equation we get,
⇒ sec 31° sin 59° + cos 31° cosec 59
⇒ 1 cos 31° × sin (90° - 31°) + cos (90° - 59°) × 1 sin 59° As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ ⇒ 1 cos 31° × cos 31° + sin 59° × 1 sin 59° ⇒ 1 + 1 ⇒ 2. \Rightarrow \dfrac{1}{\text{cos 31°}} \times \text{sin (90° - 31°)} + \text{cos (90° - 59°)} \times \dfrac{1}{\text{sin 59°}} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos 31°}} \times \text{cos 31°} + \text{sin 59°} \times \dfrac{1}{\text{sin 59°}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ cos 31° 1 × sin (90° - 31°) + cos (90° - 59°) × sin 59° 1 As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ ⇒ cos 31° 1 × cos 31° + sin 59° × sin 59° 1 ⇒ 1 + 1 ⇒ 2.
Since, L.H.S. = R.H.S.
Hence, proved that sec 31° sin 59° + cos 31° cosec 59° = 2.
Without using trigonometric tables, prove that:
sec 70° sin 20° - cos 20° cosec 70° = 0
Answer
To prove,
sec 70° sin 20° - cos 20° cosec 70° = 0
Solving L.H.S. of the equation we get :
⇒ sec 70° sin 20° - cos 20° cosec 70° ⇒ 1 cos 70° × sin (90° - 70°) − cos (90° - 70°) × 1 sin 70° As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ ⇒ 1 cos 70° × cos 70° − sin 70° × 1 sin 70° ⇒ 1 − 1 ⇒ 0. \Rightarrow \text{sec 70° sin 20° - cos 20° cosec 70°} \\[1em] \Rightarrow \dfrac{1}{\text{cos 70°}} \times \text{sin (90° - 70°)} - \text{cos (90° - 70°)} \times \dfrac{1}{\text{sin 70°}} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos 70°}} \times \text{cos 70°} - \text{sin 70°} \times \dfrac{1}{\text{sin 70°}} \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0. ⇒ sec 70° sin 20° - cos 20° cosec 70° ⇒ cos 70° 1 × sin (90° - 70°) − cos (90° - 70°) × sin 70° 1 As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ ⇒ cos 70° 1 × cos 70° − sin 70° × sin 70° 1 ⇒ 1 − 1 ⇒ 0.
Since, L.H.S. = R.H.S.
Hence, proved that sec 70° sin 20° - cos 20° cosec 70° = 0.
Without using trigonometric tables, prove that:
sin2 20° + sin2 70° - tan2 45° = 0.
Answer
To prove,
sin2 20° + sin2 70° - tan2 45° = 0.
Solving, L.H.S. of the equation we get :
sin2 20° + sin2 70° - tan2 45°
We know that,
sin (90 - θ) = cos θ
and
sin2 θ + cos2 θ = 1.
⇒ sin2 20° + sin2 (90° - 20°) - (1)2
⇒ sin2 20° + cos2 20° - 1
⇒ 1 - 1
⇒ 0.
Since, L.H.S. = R.H.S.
Hence, proved that sin2 20° + sin2 70° - tan2 45° = 0.
Without using trigonometric tables, prove that:
cot 54° tan 36° + tan 20° cot 70° − 2 = 0 \dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0 tan 36° cot 54° + cot 70° tan 20° − 2 = 0
Answer
To prove,
cot 54° tan 36° + tan 20° cot 70° − 2 = 0 \dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0 tan 36° cot 54° + cot 70° tan 20° − 2 = 0
Solving, L.H.S. of the equation we get :
⇒ cot 54° tan 36° + tan 20° cot 70° − 2 ⇒ cot (90° - 36°) tan 36° + tan 20° cot (90° - 20°) − 2 ⇒ tan 36° tan 36° + tan 20° tan 20° − 2 ⇒ 1 + 1 − 2 ⇒ 0. \Rightarrow \dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 \\[1em] \Rightarrow \dfrac{\text{cot (90° - 36°)}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot (90° - 20°)}} - 2 \\[1em] \Rightarrow \dfrac{\text{tan 36°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{tan 20°}} - 2 \\[1em] \Rightarrow 1 + 1 - 2 \\[1em] \Rightarrow 0. ⇒ tan 36° cot 54° + cot 70° tan 20° − 2 ⇒ tan 36° cot (90° - 36°) + cot (90° - 20°) tan 20° − 2 ⇒ tan 36° tan 36° + tan 20° tan 20° − 2 ⇒ 1 + 1 − 2 ⇒ 0.
Since, L.H.S. = R.H.S.
Hence, proved that cot 54° tan 36° + tan 20° cot 70° − 2 = 0 \dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0 tan 36° cot 54° + cot 70° tan 20° − 2 = 0
Without using trigonometric tables, prove that:
sin 50° cos 40° + cosec 40° sec 50° − 4 cos 50° cosec 40° + 2 = 0 \dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0 cos 40° sin 50° + sec 50° cosec 40° − 4 cos 50° cosec 40° + 2 = 0
Answer
To prove,
sin 50° cos 40° + cosec 40° sec 50° − 4 cos 50° cosec 40° + 2 = 0 \dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0 cos 40° sin 50° + sec 50° cosec 40° − 4 cos 50° cosec 40° + 2 = 0
Solving, L.H.S. of the equation we get :
⇒ sin 50° cos 40° + cosec 40° sec 50° − 4 cos 50° cosec 40° + 2 = 0. ⇒ sin (90° - 40°) cos 40° + cosec (90° - 50°) sec 50° − 4 cos 50° cosec (90° - 50°) + 2 = 0 \phantom{\Rightarrow} \dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0. \\[1em] \Rightarrow \dfrac{\text{sin (90° - 40°)}}{\text{cos 40°}} + \dfrac{\text{cosec (90° - 50°)}}{\text{sec 50°}} - \text{4 cos 50° cosec (90° - 50°) + 2} = 0 \\[1em] ⇒ cos 40° sin 50° + sec 50° cosec 40° − 4 cos 50° cosec 40° + 2 = 0. ⇒ cos 40° sin (90° - 40°) + sec 50° cosec (90° - 50°) − 4 cos 50° cosec (90° - 50°) + 2 = 0
We know that,
sin (90 - θ) = cos θ
and
cosec (90 - θ) = sec θ
⇒ cos 40° cos 40° + sec 50° sec 50° − 4 cos 50° sec 50° + 2 ⇒ 1 + 1 − 4 × cos 50° × 1 cos 50° + 2 ⇒ 2 − 4 + 2 ⇒ 0. \Rightarrow \dfrac{\text{cos 40°}}{\text{cos 40°}} + \dfrac{\text{sec 50°}}{\text{sec 50°}} - \text{4 cos 50° sec 50°} + 2 \\[1em] \Rightarrow 1 + 1 - 4 \times \text{cos 50°} \times \dfrac{1}{\text{cos 50°}} + 2 \\[1em] \Rightarrow 2 - 4 + 2 \\[1em] \Rightarrow 0. ⇒ cos 40° cos 40° + sec 50° sec 50° − 4 cos 50° sec 50° + 2 ⇒ 1 + 1 − 4 × cos 50° × cos 50° 1 + 2 ⇒ 2 − 4 + 2 ⇒ 0.
Since, L.H.S. = R.H.S.
Hence, proved that sin 50° cos 40° + cosec 40° sec 50° − 4 cos 50° cosec 40° + 2 = 0 \dfrac{\text{sin 50°}}{\text{cos 40°}} + \dfrac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0 cos 40° sin 50° + sec 50° cosec 40° − 4 cos 50° cosec 40° + 2 = 0
Without using trigonometric tables, prove that:
cos 70° sin 20° + cos 59° sin 31° − 8 sin 2 30 ° = 0 \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° = 0 sin 20° cos 70° + sin 31° cos 59° − 8 sin 2 30° = 0
Answer
To prove,
cos 70° sin 20° + cos 59° sin 31° − 8 sin 2 30 ° = 0 \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° = 0 sin 20° cos 70° + sin 31° cos 59° − 8 sin 2 30° = 0
Solving, L.H.S. of the equation we get :
⇒ cos 70° sin 20° + cos 59° sin 31° − 8 sin 2 30 ° ⇒ cos (90° - 20°) sin 20° + cos (90° - 31°) sin 31° − 8 sin 2 30 ° As, cos (90 - θ) = sin θ ⇒ sin 20° sin 20° + sin 31° sin 31° − 8 × ( 1 2 ) 2 ⇒ 1 + 1 − 2 ⇒ 0. \Rightarrow \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° \\[1em] \Rightarrow \dfrac{\text{cos (90° - 20°)}}{\text{sin 20°}} + \dfrac{\text{cos (90° - 31°)}}{\text{sin 31°}} - 8\text{ sin}^2 30° \\[1em] \text{As, cos (90 - θ) = sin θ} \\[1em] \Rightarrow \dfrac{\text{sin 20°}}{\text{sin 20°}} + \dfrac{\text{sin 31°}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2 \\[1em] \Rightarrow 1 + 1 - 2 \\[1em] \Rightarrow 0. ⇒ sin 20° cos 70° + sin 31° cos 59° − 8 sin 2 30° ⇒ sin 20° cos (90° - 20°) + sin 31° cos (90° - 31°) − 8 sin 2 30° As, cos (90 - θ) = sin θ ⇒ sin 20° sin 20° + sin 31° sin 31° − 8 × ( 2 1 ) 2 ⇒ 1 + 1 − 2 ⇒ 0.
Since, L.H.S. = R.H.S.
Hence, proved that cos 70° sin 20° + cos 59° sin 31° − 8 sin 2 30 ° = 0 \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - 8\text{ sin}^2 30° = 0 sin 20° cos 70° + sin 31° cos 59° − 8 sin 2 30° = 0
Without using trigonometric tables, prove that:
cos 80° sin 10° + cos 59° cosec 31° = 2 \dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} = 2 sin 10° cos 80° + cos 59° cosec 31° = 2
Answer
To prove,
cos 80° sin 10° + cos 59° cosec 31° = 2 \dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} = 2 sin 10° cos 80° + cos 59° cosec 31° = 2
Solving, L.H.S. of the equation we get :
⇒ cos 80° sin 10° + cos 59° cosec 31° ⇒ cos 80° sin (90° - 80°) + cos 59° cosec (90° - 59°) As, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ ⇒ cos 80° cos 80° + cos 59° sec 59° ⇒ 1 + cos 59° × 1 cos 59° ⇒ 1 + 1 ⇒ 2. \phantom{\Rightarrow} \dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} \\[1em] \Rightarrow \dfrac{\text{cos 80°}}{\text{sin (90° - 80°)}} + \text{cos 59° cosec (90° - 59°)} \\[1em] \text{As, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ } \\[1em] \Rightarrow \dfrac{\text{cos 80°}}{\text{cos 80°}} + \text{cos 59° sec 59°} \\[1em] \Rightarrow 1 + \text{cos 59°} \times \dfrac{1}{\text{cos 59°}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ sin 10° cos 80° + cos 59° cosec 31° ⇒ sin (90° - 80°) cos 80° + cos 59° cosec (90° - 59°) As, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ ⇒ cos 80° cos 80° + cos 59° sec 59° ⇒ 1 + cos 59° × cos 59° 1 ⇒ 1 + 1 ⇒ 2.
Since, L.H.S. = R.H.S.
Hence, proved that cos 80° sin 10° + cos 59° cosec 31° = 2 \dfrac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} = 2 sin 10° cos 80° + cos 59° cosec 31° = 2
Without using trigonometric tables, evaluate :
2 ( tan 35° cot 55° ) 2 + ( cot 55° tan 35° ) − 3 ( sec 40° cosec 50° ) 2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big) 2 ( cot 55° tan 35° ) 2 + ( tan 35° cot 55° ) − 3 ( cosec 50° sec 40° )
Answer
Solving,
2 ( tan 35° cot 55° ) 2 + ( cot 55° tan 35° ) − 3 ( sec 40° cosec 50° ) ⇒ 2 ( tan 35° cot (90° - 35°) ) 2 + ( cot (90° - 35°) tan 35° ) − 3 ( sec (90° - 50°) cosec 50° ) As, cot (90 - θ) = tan θ and sec (90 - θ) = cosec θ ⇒ 2 ( tan 35° tan 35° ) 2 + ( tan 35° tan 35° ) − 3 ( cosec 50° cosec 50° ) ⇒ 2 ( 1 ) 2 + 1 − 3 ⇒ 3 − 3 ⇒ 0. 2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)\\[1em] \Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{cot (90° - 35°)}}\Big)^2 + \Big(\dfrac{\text{cot (90° - 35°)}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec (90° - 50°)}}{\text{cosec 50°}}\Big)\\[1em] \text{As, cot (90 - θ) = tan θ and sec (90 - θ) = cosec θ} \\[1em] \Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 + \Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{cosec 50°}}{\text{cosec 50°}}\Big)\\[1em] \Rightarrow 2(1)^2 + 1 - 3 \\[1em] \Rightarrow 3 - 3 \\[1em] \Rightarrow 0. 2 ( cot 55° tan 35° ) 2 + ( tan 35° cot 55° ) − 3 ( cosec 50° sec 40° ) ⇒ 2 ( cot (90° - 35°) tan 35° ) 2 + ( tan 35° cot (90° - 35°) ) − 3 ( cosec 50° sec (90° - 50°) ) As, cot (90 - θ) = tan θ and sec (90 - θ) = cosec θ ⇒ 2 ( tan 35° tan 35° ) 2 + ( tan 35° tan 35° ) − 3 ( cosec 50° cosec 50° ) ⇒ 2 ( 1 ) 2 + 1 − 3 ⇒ 3 − 3 ⇒ 0.
Hence, 2 ( tan 35° cot 55° ) 2 + ( cot 55° tan 35° ) − 3 ( sec 40° cosec 50° ) = 0 2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big) = 0 2 ( cot 55° tan 35° ) 2 + ( tan 35° cot 55° ) − 3 ( cosec 50° sec 40° ) = 0
Without using trigonometric tables, evaluate :
( sin 35° cos 55° + cos 35° sin 55° cosec 2 10 ° − tan 2 80 ° ) . \Big(\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}\Big). ( cosec 2 10° − tan 2 80° sin 35° cos 55° + cos 35° sin 55° ) .
Answer
Solving,
⇒ ( sin 35° cos 55° + cos 35° sin 55° cosec 2 10 ° − tan 2 80 ° ) ⇒ ( sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°) cosec 2 ( 90 ° − 80 ° ) − tan 2 80 ° ) As, sin (90 - θ) = cos θ, cos (90 - θ) = sin θ and cosec (90 - θ) = sec θ ⇒ ( sin 35° sin 35° + cos 35° cos 35° sec 2 80 ° − tan 2 80 ° ) ⇒ ( sin 2 35 ° + cos 2 35 ° sec 2 80 ° − tan 2 80 ° ) As, sin 2 θ + cos 2 θ = 1 and sec 2 θ − tan 2 θ = 1 ⇒ 1 1 ⇒ 1. \Rightarrow \Big(\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 \space 10° - \text{tan}^2 \space 80°}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°)}}{\text{cosec}^2 \space (90° - 80°) - \text{tan}^2 \space 80°}\Big) \\[1em] \text{As, sin (90 - θ) = cos θ, cos (90 - θ) = sin θ and cosec (90 - θ) = sec θ} \\[1em] \Rightarrow \Big(\dfrac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{sec}^2 \space 80° - \text{tan}^2 \space 80°}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin}^2 \space 35° + \text{cos}^2 \space 35°}{\text{sec}^2 \space 80° - \text{tan}^2 \space 80°}\Big) \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \text{ and } \text{sec}^2 θ - \text{tan}^2 θ = 1 \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1. ⇒ ( cosec 2 10° − tan 2 80° sin 35° cos 55° + cos 35° sin 55° ) ⇒ ( cosec 2 ( 90° − 80° ) − tan 2 80° sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°) ) As, sin (90 - θ) = cos θ, cos (90 - θ) = sin θ and cosec (90 - θ) = sec θ ⇒ ( sec 2 80° − tan 2 80° sin 35° sin 35° + cos 35° cos 35° ) ⇒ ( sec 2 80° − tan 2 80° sin 2 35° + cos 2 35° ) As, sin 2 θ + cos 2 θ = 1 and sec 2 θ − tan 2 θ = 1 ⇒ 1 1 ⇒ 1.
Hence, ( sin 35° cos 55° + cos 35° sin 55° cosec 2 10 ° − tan 2 80 ° ) \Big(\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 \space 10° - \text{tan}^2 \space 80°}\Big) ( cosec 2 10° − tan 2 80° sin 35° cos 55° + cos 35° sin 55° )
Without using trigonometric tables, evaluate :
sin2 34° + sin2 56° + 2 tan 18° tan 72° - cot2 30°.
Answer
Solving,
⇒ sin2 34° + sin2 56° + 2 tan 18° tan 72° - cot2 30°
⇒ sin2 34° + sin2 (90° - 34°) + 2 tan (90° - 72°) tan 72° - cot2 30°
We know that,
sin (90° - θ) = cos θ, cos (90° - θ) = sin θ and tan (90° - θ) = cot θ
⇒ sin2 34° + cos2 34° + 2 cot 72° tan 72° - cot2 30°
As,
cot θ. tan θ = 1 and sin2 θ + cos2 θ = 1.
⇒ 1 + 2 - ( 3 ) 2 (\sqrt{3})^2 ( 3 ) 2
⇒ 3 - 3
⇒ 0.
Hence, sin2 34° + sin2 56° + 2 tan 18° tan 72° - cot2 30 = 0.
Prove the following :
cos θ sin (90° - θ) + sin θ cos (90° - θ) = 2 \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{sin θ}}{\text{cos (90° - θ)}} = 2 sin (90° - θ) cos θ + cos (90° - θ) sin θ = 2
Answer
To prove,
cos θ sin (90° - θ) + sin θ cos (90° - θ) = 2 \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{sin θ}}{\text{cos (90° - θ)}} = 2 sin (90° - θ) cos θ + cos (90° - θ) sin θ = 2
We know that,
sin (90 - θ) = cos θ and cos (90 - θ) = sin θ
Solving L.H.S. of the equation, we get :
⇒ cos θ cos θ + sin θ sin θ ⇒ 1 + 1 ⇒ 2. \Rightarrow \dfrac{\text{cos θ}}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{sin θ}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ cos θ cos θ + sin θ sin θ ⇒ 1 + 1 ⇒ 2.
Since, L.H.S. = R.H.S.
Hence, proved that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 2 \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{sin θ}}{\text{cos (90° - θ)}} = 2 sin (90° - θ) cos θ + cos (90° - θ) sin θ = 2
Prove the following :
cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1
Answer
To prove,
cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
We know that,
sin (90 - θ) = cos θ and cos (90 - θ) = sin θ
Solving L.H.S. of the equation, we get :
⇒ cos θ cos θ + sin θ sin θ
⇒ cos2 θ + sin2 θ
⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove the following :
tan θ tan (90° - θ) + sin (90° - θ) cos θ = sec 2 θ \dfrac{\text{tan θ}}{\text{\text{tan (90° - θ)}}} + \dfrac{\text{sin (90° - θ)}}{\text{cos θ}} = \text{sec}^2 \text{ θ} tan (90° - θ) tan θ + cos θ sin (90° - θ) = sec 2 θ
Answer
To prove,
tan θ tan (90° - θ) + sin (90° - θ) cos θ = sec 2 θ . \dfrac{\text{tan θ}}{\text{\text{tan (90° - θ)}}} + \dfrac{\text{sin (90° - θ)}}{\text{cos θ}} = \text{sec}^2 \text{ θ}. tan (90° - θ) tan θ + cos θ sin (90° - θ) = sec 2 θ .
We know that,
sin (90 - θ) = cos θ and tan (90 - θ) = cot θ
Solving L.H.S. of the equation, we get :
⇒ tan θ cot θ + cos θ cos θ ⇒ tan θ 1 tan θ + 1 ⇒ tan 2 θ + 1 ⇒ sec 2 θ . \Rightarrow \dfrac{\text{tan θ}}{\text{\text{cot θ}}} + \dfrac{\text{cos θ}}{\text{cos θ}} \\[1em] \Rightarrow \dfrac{\text{tan θ}}{\dfrac{1}{\text{tan θ}}} + 1 \\[1em] \Rightarrow \text{tan}^2 \text{ θ} + 1 \\[1em] \Rightarrow \text{sec}^2 \text{ θ}. ⇒ cot θ tan θ + cos θ cos θ ⇒ tan θ 1 tan θ + 1 ⇒ tan 2 θ + 1 ⇒ sec 2 θ .
Since, L.H.S. = R.H.S.
Hence, proved that tan θ tan (90° - θ) + sin (90° - θ) cos θ = sec 2 θ . \dfrac{\text{tan θ}}{\text{\text{tan (90° - θ)}}} + \dfrac{\text{sin (90° - θ)}}{\text{cos θ}} = \text{sec}^2 \text{ θ}. tan (90° - θ) tan θ + cos θ sin (90° - θ) = sec 2 θ .
Prove the following :
cos (90° - A) sin (90° - A) tan ( 90 ° − A ) = 1 − cos 2 A \dfrac{\text{cos (90° - A) sin (90° - A)}}{\text{tan} (90° - \text{ A})} = 1 - \text{cos}^ 2 \text{ A} tan ( 90° − A ) cos (90° - A) sin (90° - A) = 1 − cos 2 A
Answer
We know that,
cos (90 - θ) = sin θ, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ.
Solving L.H.S. of the equation, we get :
⇒ sin A cos A cot A ⇒ sin A cos A cos A sin A ⇒ sin 2 A ⇒ 1 − cos 2 A . \Rightarrow \dfrac{\text{sin A cos A}}{\text{cot A}} \\[1em] \Rightarrow \dfrac{\text{sin A cos A}}{\dfrac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \text{sin}^2 \text{ A} \\[1em] \Rightarrow 1 - \text{cos}^2 \text{ A}. ⇒ cot A sin A cos A ⇒ sin A cos A sin A cos A ⇒ sin 2 A ⇒ 1 − cos 2 A .
Since, L.H.S. = R.H.S.
Hence, proved that cos (90° - A) sin (90° - A) tan ( 90 ° − A ) = 1 − cos 2 A \dfrac{\text{cos (90° - A) sin (90° - A)}}{\text{tan} (90° - \text{ A})} = 1 - \text{cos}^ 2 \text{ A} tan ( 90° − A ) cos (90° - A) sin (90° - A) = 1 − cos 2 A
Prove the following :
sin (90° - A) cosec (90° - A) + cos (90° - A) sec (90° - A) \dfrac{\text{sin (90° - A)}}{\text{cosec (90° - A)}} + \dfrac{\text{cos (90° - A)}}{\text{sec (90° - A)}} cosec (90° - A) sin (90° - A) + sec (90° - A) cos (90° - A)
Answer
We know that,
cos (90 - θ) = sin θ, sec (90 - θ) = cosec θ, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ.
Solving L.H.S. of the equation, we get :
⇒ cos A sec A + sin A cosec A ⇒ cos A 1 cos A + sin A 1 sin A ⇒ cos 2 A + sin 2 A ⇒ 1. \Rightarrow \dfrac{\text{cos A}}{\text{sec A}} + \dfrac{\text{sin A}}{\text{cosec A}} \\[1em] \Rightarrow \dfrac{\text{cos A}}{\dfrac{1}{\text{cos A}}} + \dfrac{\text{sin A}}{\dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \text{cos}^2 A + \text{sin}^2 A \\[1em] \Rightarrow 1. ⇒ sec A cos A + cosec A sin A ⇒ cos A 1 cos A + sin A 1 sin A ⇒ cos 2 A + sin 2 A ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that sin (90° - A) cosec (90° - A) + cos (90° - A) sec (90° - A) \dfrac{\text{sin (90° - A)}}{\text{cosec (90° - A)}} + \dfrac{\text{cos (90° - A)}}{\text{sec (90° - A)}} cosec (90° - A) sin (90° - A) + sec (90° - A) cos (90° - A)
Simplify the following :
cos θ sin (90° - θ) + cos (90° - θ) sec (90° - θ) − 3 tan 2 30 ° \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30° sin (90° - θ) cos θ + sec (90° - θ) cos (90° - θ) − 3 tan 2 30°
Answer
We know that,
sin (90° - θ) = cos θ, cos (90° - θ) = sin θ and sec (90° - θ) = cosec θ.
Substituting values in equation, we get :
⇒ cos θ sin (90° - θ) + cos (90° - θ) sec (90° - θ) − 3 tan 2 30 ° ⇒ cos θ cos θ + sin θ cosec θ − 3 tan 2 30 ° ⇒ 1 + sin θ 1 sin θ − 3 × ( 1 3 ) 2 = 1 + sin 2 θ − 3 × 1 3 = 1 − 1 + sin 2 θ = sin 2 θ . \Rightarrow \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30° \\[1em] \Rightarrow \dfrac{\text{cos θ}}{\text{cos θ}} + \dfrac{\text{\text{sin θ}}}{\text{cosec θ}} - \text{3 tan}^2 30° \\[1em] \Rightarrow 1 + \dfrac{\text{sin θ}}{\dfrac{1}{\text{sin θ}}} -3 \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] = 1 + \text{sin}^2 θ - 3 \times \dfrac{1}{3} \\[1em] = 1 - 1 + \text{sin}^2 θ\\[1em] = \text{sin}^2 θ. ⇒ sin (90° - θ) cos θ + sec (90° - θ) cos (90° - θ) − 3 tan 2 30° ⇒ cos θ cos θ + cosec θ sin θ − 3 tan 2 30° ⇒ 1 + sin θ 1 sin θ − 3 × ( 3 1 ) 2 = 1 + sin 2 θ − 3 × 3 1 = 1 − 1 + sin 2 θ = sin 2 θ .
Hence, cos θ sin (90° - θ) + cos (90° - θ) sec (90° - θ) − 3 tan 2 30 ° \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30° sin (90° - θ) cos θ + sec (90° - θ) cos (90° - θ) − 3 tan 2 30° 2 θ.
Simplify the following :
cosec (90° - θ) sin (90° - θ) cot(90° - θ) cos (90° - θ) sec (90° - θ) tan θ + cot θ tan (90° - θ) . \dfrac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \dfrac{\text{cot θ}}{\text{tan (90° - θ)}}. cos (90° - θ) sec (90° - θ) tan θ cosec (90° - θ) sin (90° - θ) cot(90° - θ) + tan (90° - θ) cot θ .
Answer
We know that,
sin (90° - θ) = cos θ
Substituting values in equation, we get :
⇒ cosec (90° - θ) sin (90° - θ) cot(90° - θ) cos (90° - θ) sec (90° - θ) tan θ + cot θ tan (90° - θ) ⇒ sec θ cos θ tan θ sin θ cosec θ tan θ + cot θ cot θ ⇒ 1 cos θ × cos θ × tan θ sin θ × 1 sin θ × tan θ + 1 ⇒ 1 + 1 ⇒ 2. \Rightarrow \dfrac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \dfrac{\text{cot θ}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \dfrac{\text{sec θ cos θ tan θ}}{\text{sin θ cosec θ tan θ}} + \dfrac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos θ}} \times \text{cos θ} \times \text{tan θ}}{\text{sin θ} \times \dfrac{1}{\text{sin θ}} \times \text{tan θ}} + 1 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ cos (90° - θ) sec (90° - θ) tan θ cosec (90° - θ) sin (90° - θ) cot(90° - θ) + tan (90° - θ) cot θ ⇒ sin θ cosec θ tan θ sec θ cos θ tan θ + cot θ cot θ ⇒ sin θ × sin θ 1 × tan θ cos θ 1 × cos θ × tan θ + 1 ⇒ 1 + 1 ⇒ 2.
Hence, cosec (90° - θ) sin (90° - θ) cot(90° - θ) cos (90° - θ) sec (90° - θ) tan θ + cot θ tan (90° - θ) \dfrac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \dfrac{\text{cot θ}}{\text{tan (90° - θ)}} cos (90° - θ) sec (90° - θ) tan θ cosec (90° - θ) sin (90° - θ) cot(90° - θ) + tan (90° - θ) cot θ
Show that :
cos 2 ( 45 ° + θ ) + cos 2 ( 45 ° − θ ) tan (60° + θ) tan (30° - θ) = 1 \dfrac{\text{cos}^2 (45° + θ) + \text{cos}^2 (45° - θ)}{\text{tan (60° + θ) \text{tan (30° - θ)}}} = 1 tan (60° + θ) tan (30° - θ) cos 2 ( 45° + θ ) + cos 2 ( 45° − θ ) = 1
Answer
Solving, L.H.S. of the equation, we get :
⇒ cos 2 ( 45 ° + θ ) + cos 2 ( 45 ° − θ ) tan (60° + θ) tan (30° - θ) ⇒ cos 2 ( 45 ° + θ ) + sin 2 [ 90 ° − ( 45 ° − θ ) ] tan (60° + θ) cot [90° - (30° - θ)] ⇒ cos 2 ( 45 ° + θ ) + sin 2 ( 45 ° + θ ) tan (60° + θ) cot (60° + θ) As, cos 2 A + sin 2 A = 1 and tan A. cot A = 1 ⇒ 1 1 ⇒ 1. \Rightarrow \dfrac{\text{cos}^2 (45° + θ) + \text{cos}^2 (45° - θ)}{\text{tan (60° + θ) \text{tan (30° - θ)}}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 (45° + θ) + \text{sin}^2 [90° - (45° - θ)]}{\text{tan (60° + θ) \text{cot [90° - (30° - θ)]}}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 (45° + θ) + \text{sin}^2 (45° + θ)}{\text{tan (60° + θ) \text{cot (60° + θ)}}} \\[1em] \text{As, cos}^2 \text{ A + sin}^2 A = 1 \text{ and tan A. cot A} = 1 \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1. ⇒ tan (60° + θ) tan (30° - θ) cos 2 ( 45° + θ ) + cos 2 ( 45° − θ ) ⇒ tan (60° + θ) cot [90° - (30° - θ)] cos 2 ( 45° + θ ) + sin 2 [ 90° − ( 45° − θ )] ⇒ tan (60° + θ) cot (60° + θ) cos 2 ( 45° + θ ) + sin 2 ( 45° + θ ) As, cos 2 A + sin 2 A = 1 and tan A. cot A = 1 ⇒ 1 1 ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that cos 2 ( 45 ° + θ ) + cos 2 ( 45 ° − θ ) tan (60° + θ) tan (30° - θ) = 1 \dfrac{\text{cos}^2 (45° + θ) + \text{cos}^2 (45° - θ)}{\text{tan (60° + θ) \text{tan (30° - θ)}}} = 1 tan (60° + θ) tan (30° - θ) cos 2 ( 45° + θ ) + cos 2 ( 45° − θ ) = 1
Find the value of A if
sin 3A = cos (A - 6°), where 3A and A - 6° are acute angles.
Answer
Given,
sin 3A = cos (A - 6°)
⇒ sin 3A = sin [90° - (A - 6°)]
⇒ 3A = 90° - (A - 6°)
⇒ 3A = 96° - A
⇒ 4A = 96°
⇒ A = 96 ° 4 \dfrac{96°}{4} 4 96°
⇒ A = 24°.
Hence, A = 24°.
Find the value of A if
tan 2A = cot (A - 18°), where 2A and A - 18° are acute angles.
Answer
Given,
tan 2A = cot (A - 18°)
⇒ tan 2A = tan [90° - (A - 18°)]
⇒ 2A = 90° - (A - 18°)
⇒ 2A = 90° - A + 18°
⇒ 2A = 108° - A
⇒ 3A = 108°
⇒ 3A = 108 ° 3 \dfrac{108°}{3} 3 108°
⇒ A = 36°
Hence, A = 36°.
Find the value of A if
If sec 2A = cosec (A - 27°) where 2A is an acute angle, find the measure of ∠A.
Answer
Given,
sec 2A = cosec (A - 27°)
⇒ sec 2A = sec [90° - (A - 27°)]
⇒ 2A = 90° - (A - 27°)
⇒ 2A = 90° - A + 27°
⇒ 2A = 117° - A
⇒ 3A = 117°
⇒ A = 117 ° 3 \dfrac{117°}{3} 3 117°
⇒ A = 39°.
Hence, A = 39°.
Find the value of θ (0° < θ < 90°) if :
cos 63° sec (90° - θ) = 1
Answer
Given,
cos 63° sec (90° - θ) = 1
We know that,
cos A sec A = 1
∴ 90° - θ = 63°
⇒ θ = 90° - 63°
⇒ θ = 27°.
Hence, θ = 27°.
Find the value of θ (0° < θ < 90°) if :
tan 35° cot (90° - θ) = 1.
Answer
Given,
tan 35° cot (90° - θ) = 1
We know that,
tan A cot A = 1
∴ 90° - θ = 35°
⇒ θ = 90° - 35°
⇒ θ = 55°.
Hence, θ = 55°.
If A, B and C are the interior angles of a △ABC, show that :
(i) cos A + B 2 \dfrac{A + B}{2} 2 A + B C 2 \dfrac{C}{2} 2 C
(ii) tan C + A 2 \dfrac{C + A}{2} 2 C + A B 2 \dfrac{B}{2} 2 B
Answer
(i) Given,
A, B and C are the interior angles of a △ABC.
∴ A + B + C = 180°
⇒ A + B + C 2 = 90 ° \dfrac{A + B + C}{2} = 90° 2 A + B + C = 90°
⇒ A + B 2 = 90 ° − C 2 \dfrac{A + B}{2} = 90° - \dfrac{C}{2} 2 A + B = 90° − 2 C
To prove,
cos A + B 2 \dfrac{A + B}{2} 2 A + B C 2 \dfrac{C}{2} 2 C
Substituting value of A + B 2 \dfrac{A + B}{2} 2 A + B
= cos ( A + B 2 ) = cos ( 90 ° − C 2 ) = sin C 2 [ ∵ cos (90 - θ) = sin θ ] \phantom{=} \text{cos } \Big(\dfrac{A + B}{2}\Big) \\[1em] = \text{cos } \Big(90\degree - \dfrac{C}{2}\Big) \\[1em] = \text{sin } \dfrac{C}{2} \space [\because \text{cos (90 - θ)} = \text{sin θ}] = cos ( 2 A + B ) = cos ( 90° − 2 C ) = sin 2 C [ ∵ cos (90 - θ) = sin θ ]
Since, L.H.S. = R.H.S.
Hence, proved that cos A + B 2 \dfrac{A + B}{2} 2 A + B C 2 \dfrac{C}{2} 2 C
(ii) Given,
A, B and C are the interior angles of a △ABC.
∴ A + B + C = 180°
⇒ A + B + C 2 = 90 ° \dfrac{A + B + C}{2} = 90° 2 A + B + C = 90°
⇒ A + C 2 = 90 ° − B 2 \dfrac{A + C}{2} = 90° - \dfrac{B}{2} 2 A + C = 90° − 2 B
To prove,
tan C + A 2 \dfrac{C + A}{2} 2 C + A B 2 \dfrac{B}{2} 2 B
Substituting value of C + A 2 \dfrac{C + A}{2} 2 C + A
= tan ( C + A 2 ) = tan ( 90 ° − B 2 ) = cot B 2 [ ∵ tan (90 - θ) = cot θ ] \phantom{=} \text{tan } \Big(\dfrac{C + A}{2}\Big) \\[1em] = \text{tan } \Big(90\degree - \dfrac{B}{2}\Big) \\[1em] = \text{cot } \dfrac{B}{2} \space [\because \text{tan (90 - θ)} = \text{cot θ}] = tan ( 2 C + A ) = tan ( 90° − 2 B ) = cot 2 B [ ∵ tan (90 - θ) = cot θ ]
Since, L.H.S. = R.H.S.
Hence, proved that tan C + A 2 \dfrac{C + A}{2} 2 C + A B 2 \dfrac{B}{2} 2 B
