In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C
Answer
ABC is a right angled triangle as shown below:
By pythagoras theorem,
⇒ AC2 = AB2 + BC2
⇒ AC2 = 242 + 72
⇒ AC2 = 576 + 49
⇒ AC2 = 625
⇒ AC = 625 \sqrt{625} 625
(i) sin A = Side opposite to ∠A Hypotenuse = B C A C \dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} Hypotenuse Side opposite to ∠A = A C BC
Substituting values we get,
sin A = 7 25 \dfrac{7}{25} 25 7
cos A = Side adjacent to ∠A Hypotenuse = A B A C \dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} Hypotenuse Side adjacent to ∠A = A C A B
Substituting values we get,
cos A = 24 25 \dfrac{24}{25} 25 24
Hence, sin A = 7 25 \dfrac{7}{25} 25 7 24 25 \dfrac{24}{25} 25 24
(ii) sin C = Side opposite to ∠C Hypotenuse = A B A C \dfrac{\text{Side opposite to ∠C}}{\text{Hypotenuse}} = \dfrac{AB}{AC} Hypotenuse Side opposite to ∠C = A C A B
Substituting values we get,
sin C = 24 25 \dfrac{24}{25} 25 24
cos C = Side adjacent to ∠C Hypotenuse = B C A C \dfrac{\text{Side adjacent to ∠C}}{\text{Hypotenuse}} = \dfrac{BC}{AC} Hypotenuse Side adjacent to ∠C = A C BC
Substituting values we get,
cos C = 7 25 \dfrac{7}{25} 25 7
Hence, sin C = 24 25 \dfrac{24}{25} 25 24 7 25 \dfrac{7}{25} 25 7
In the given figure, find tan P – cot R.
Answer
From figure,
PQR is a right angled triangle.
By pythagoras theorem, we get :
⇒ PR2 = PQ2 + QR2
⇒ 132 = 122 + QR2
⇒ QR2 = 169 - 144
⇒ QR2 = 25
⇒ QR = 25 \sqrt{25} 25
tan P = Side opposite to ∠P Side adjacent to ∠P = Q R P Q = 5 12 \dfrac{\text{Side opposite to ∠P}}{\text{Side adjacent to ∠P}} = \dfrac{QR}{PQ} = \dfrac{5}{12} Side adjacent to ∠P Side opposite to ∠P = PQ QR = 12 5
cot R = Side adjacent to ∠R Side opposite to ∠R = Q R P Q = 5 12 \dfrac{\text{Side adjacent to ∠R}}{\text{Side opposite to ∠R}} = \dfrac{QR}{PQ} = \dfrac{5}{12} Side opposite to ∠R Side adjacent to ∠R = PQ QR = 12 5
Substituting values in tan P - cot R, we get :
⇒ 5 12 − 5 12 \Rightarrow \dfrac{5}{12} - \dfrac{5}{12} ⇒ 12 5 − 12 5
Hence, tan P - cot R = 0.
If sin A = 3 4 \dfrac{3}{4} 4 3
Answer
Let us draw a right angle triangle ABC.
We know that,
sin A = Side opposite to ∠A Hypotenuse \dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} Hypotenuse Side opposite to ∠A
Substituting values we get,
3 4 = B C A C \dfrac{3}{4} = \dfrac{BC}{AC} 4 3 = A C BC
Let BC = 3k and AC = 4k.
In right angle triangle ABC,
By pythagoras theorem,
⇒ AC2 = AB2 + BC2
⇒ (4k)2 = AB2 + (3k)2
⇒ AB2 = 16k2 - 9k2
⇒ AB2 = 7k2
⇒ AB = 7 k 2 = 7 k \sqrt{7k^2} = \sqrt{7}k 7 k 2 = 7 k
We know that,
cos A = Side adjacent to ∠A Hypotenuse = A B A C = 7 k 4 k = 7 4 \dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{7}k}{4k} = \dfrac{\sqrt{7}}{4} Hypotenuse Side adjacent to ∠A = A C A B = 4 k 7 k = 4 7
tan A = Side opposite to ∠A Side adjacent to ∠A = B C A B = 3 k 7 k = 3 7 \dfrac{\text{Side opposite to ∠A}}{\text{Side adjacent to ∠A}} = \dfrac{BC}{AB} = \dfrac{3k}{\sqrt{7}k} = \dfrac{3}{\sqrt{7}} Side adjacent to ∠A Side opposite to ∠A = A B BC = 7 k 3 k = 7 3
Hence, cos A = 7 4 and tan A = 3 7 \text{cos A} = \dfrac{\sqrt{7}}{4} \text{ and tan A} = \dfrac{3}{\sqrt{7}} cos A = 4 7 and tan A = 7 3
Given 15 cot A = 8, find sin A and sec A.
Answer
Given,
⇒ 15 cot A = 8
⇒ cot A = 8 15 \dfrac{8}{15} 15 8
Let us draw a right angle triangle ABC.
We know that,
cot A = Side adjacent to ∠A Side opposite to ∠A \dfrac{\text{Side adjacent to ∠A}}{\text{Side opposite to ∠A}} Side opposite to ∠A Side adjacent to ∠A
Substituting values we get,
8 15 = A B B C \dfrac{8}{15} = \dfrac{AB}{BC} 15 8 = BC A B
Let AB = 8k and BC = 15k.
By pythagoras theorem,
⇒ AC2 = AB2 + BC2
⇒ AC2 = (8k)2 + (15k)2
⇒ AC2 = 64k2 + 225k2
⇒ AC2 = 289k2
⇒ AC = 289 k 2 \sqrt{289k^2} 289 k 2
We know that,
sin A = Side opposite to ∠A Hypotenuse = B C A C = 15 k 17 k = 15 17 \dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{15k}{17k} = \dfrac{15}{17} Hypotenuse Side opposite to ∠A = A C BC = 17 k 15 k = 17 15
sec A = Hypotenuse Side adjacent to ∠A = A C A B = 17 k 8 k = 17 8 \dfrac{\text{Hypotenuse}}{\text{Side adjacent to ∠A}} = \dfrac{AC}{AB} = \dfrac{17k}{8k} = \dfrac{17}{8} Side adjacent to ∠A Hypotenuse = A B A C = 8 k 17 k = 8 17
Hence, sin A = 15 17 and sec A = 17 8 \text{sin A} = \dfrac{15}{17} \text{ and sec A} = \dfrac{17}{8} sin A = 17 15 and sec A = 8 17
Given sec θ = 13 12 \dfrac{13}{12} 12 13
Answer
Let us draw a right angle triangle ABC, with ∠A = θ.
We know that,
sec θ = Hypotenuse Side adjacent to angle θ \dfrac{\text{Hypotenuse}}{\text{Side adjacent to angle θ}} Side adjacent to angle θ Hypotenuse
Substituting values we get,
13 12 = A C A B \dfrac{13}{12} = \dfrac{AC}{AB} 12 13 = A B A C
Let AC = 13k and AB = 12k.
In right angle triangle ABC,
By pythagoras theorem,
⇒ AC2 = AB2 + BC2
⇒ (13k)2 = (12k)2 + BC2
⇒ BC2 = 169k2 - 144k2
⇒ BC2 = 25k2
⇒ BC = 25 k 2 \sqrt{25k^2} 25 k 2
We know that,
⇒ sin θ = Side opposite to angle θ Hypotenuse = B C A C = 5 k 13 k = 5 13 \dfrac{\text{Side opposite to angle θ}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{5k}{13k} = \dfrac{5}{13} Hypotenuse Side opposite to angle θ = A C BC = 13 k 5 k = 13 5
⇒ cos θ = 1 sec θ = 1 13 12 = 12 13 \dfrac{1}{\text{sec θ}} = \dfrac{1}{\dfrac{13}{12}} = \dfrac{12}{13} sec θ 1 = 12 13 1 = 13 12
⇒ tan θ = Side opposite to angle θ Side adjacent to angle θ = B C A B = 5 k 12 k = 5 12 \dfrac{\text{Side opposite to angle θ}}{\text{Side adjacent to angle θ}} = \dfrac{BC}{AB} = \dfrac{5k}{12k} = \dfrac{5}{12} Side adjacent to angle θ Side opposite to angle θ = A B BC = 12 k 5 k = 12 5
⇒ cot θ = 1 tan θ = 1 5 12 = 12 5 \dfrac{1}{\text{tan θ}} = \dfrac{1}{\dfrac{5}{12}} = \dfrac{12}{5} tan θ 1 = 12 5 1 = 5 12
⇒ cosec θ = 1 sin θ = 1 5 13 = 13 5 \dfrac{1}{\text{sin θ}} = \dfrac{1}{\dfrac{5}{13}} = \dfrac{13}{5} sin θ 1 = 13 5 1 = 5 13
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer
Let us consider two right triangles ACD and BEF where cos A = cos B.
We know that,
cos A = Side adjacent to angle A Hypotenuse = A C A D \dfrac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \dfrac{AC}{AD} Hypotenuse Side adjacent to angle A = A D A C
cos B = Side adjacent to angle B Hypotenuse = B E B F \dfrac{\text{Side adjacent to angle B}}{\text{Hypotenuse}} = \dfrac{BE}{BF} Hypotenuse Side adjacent to angle B = BF BE
Given,
⇒ cos A = cos B
⇒ A C A D = B E B F \dfrac{AC}{AD} = \dfrac{BE}{BF} A D A C = BF BE
⇒ A C B E = A D B F \dfrac{AC}{BE} = \dfrac{AD}{BF} BE A C = BF A D
⇒ AC = k.BE and AD = k.BF
In right angle triangle ACD,
By pythagoras theorem,
⇒ AD2 = AC2 + CD2
⇒ CD2 = AD2 - AC2
⇒ CD2 = (k.BF)2 - (k.BE)2
⇒ CD2 = k2 (BF2 - BE2 )
⇒ CD = k 2 ( B F 2 − B E 2 ) = k ( B F 2 − B E 2 ) \sqrt{k^2(BF^2 - BE^2)} = k\sqrt{(BF^2 - BE^2)} k 2 ( B F 2 − B E 2 ) = k ( B F 2 − B E 2 )
In right angle triangle BEF,
By pythagoras theorem,
⇒ BF2 = BE2 + EF2
⇒ EF2 = BF2 - BE2
⇒ EF = B F 2 − B E 2 \sqrt{BF^2 - BE^2} B F 2 − B E 2
So,
C D E F = k B F 2 − B E 2 B F 2 − B E 2 \dfrac{CD}{EF} = \dfrac{k\sqrt{BF^2 - BE^2}}{\sqrt{BF^2 - BE^2}} EF C D = B F 2 − B E 2 k B F 2 − B E 2
From (1) and (2), we get :
A C B E = A D B F = C D F E \dfrac{AC}{BE} = \dfrac{AD}{BF} = \dfrac{CD}{FE} BE A C = BF A D = FE C D
Since, ratio of corresponding sides of similar triangle are proportional.
∴ △ACD ~ △BEF.
∴ ∠A = ∠B.
Hence, proved that ∠A = ∠B.
If cot θ = 7 8 \dfrac{7}{8} 8 7
(i) (1 + sin θ)(1 - sin θ) (1 + cos θ)(1 - cos θ) \dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}} (1 + cos θ)(1 - cos θ) (1 + sin θ)(1 - sin θ)
(ii) cot2 θ
Answer
Let us draw a right angle triangle ABC, with ∠A = θ.
We know that,
cot θ = Side adjacent to ∠θ Side opposite to ∠θ \dfrac{\text{Side adjacent to ∠θ}}{\text{Side opposite to ∠θ}} Side opposite to ∠θ Side adjacent to ∠θ
Substituting values we get :
7 8 = A B B C \dfrac{7}{8} = \dfrac{AB}{BC} 8 7 = BC A B
Let AB = 7k and BC = 8k.
In right angle triangle ABC,
⇒ AC2 = AB2 + BC 2
⇒ AC2 = (7k)2 + (8k)2
⇒ AC2 = 49k2 + 64k2
⇒ AC2 = 113k2
⇒ AC = 113 k \sqrt{113}k 113 k
(i) We know that,
sin θ = Side opposite to ∠θ Hypotenuse = B C A C = 8 k 113 k = 8 113 \dfrac{\text{Side opposite to ∠θ}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{8k}{\sqrt{113}k} = \dfrac{8}{\sqrt{113}} Hypotenuse Side opposite to ∠θ = A C BC = 113 k 8 k = 113 8
cos θ = Side adjacent to ∠θ Hypotenuse = A B A C = 7 k 113 k = 7 113 \dfrac{\text{Side adjacent to ∠θ}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{7k}{\sqrt{113}k} = \dfrac{7}{\sqrt{113}} Hypotenuse Side adjacent to ∠θ = A C A B = 113 k 7 k = 113 7
Substituting values of sin θ and cos θ in (1 + sin θ)(1 - sin θ) (1 + cos θ)(1 - cos θ) \dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}} (1 + cos θ)(1 - cos θ) (1 + sin θ)(1 - sin θ)
⇒ ( 1 + 8 113 ) ( 1 − 8 113 ) ( 1 + 7 113 ) ( 1 − 7 113 ) ⇒ ( 1 ) 2 − ( 8 113 ) 2 ( 1 ) 2 − ( 7 113 ) 2 ⇒ 1 − 64 113 1 − 49 113 ⇒ 113 − 64 113 113 − 49 113 ⇒ 49 113 64 113 ⇒ 49 64 . \Rightarrow \dfrac{\Big(1 + \dfrac{8}{\sqrt{113}}\Big)\Big(1 - \dfrac{8}{\sqrt{113}}\Big)}{\Big(1 + \dfrac{7}{\sqrt{113}}\Big)\Big(1 - \dfrac{7}{\sqrt{113}}\Big)} \\[1em] \Rightarrow \dfrac{(1)^2 - \Big(\dfrac{8}{\sqrt{113}}\Big)^2}{(1)^2 - \Big(\dfrac{7}{\sqrt{113}}\Big)^2} \\[1em] \Rightarrow \dfrac{1 - \dfrac{64}{113}}{1 - \dfrac{49}{113}} \\[1em] \Rightarrow \dfrac{\dfrac{113 - 64}{113}}{\dfrac{113 - 49}{113}} \\[1em] \Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}} \\[1em] \Rightarrow \dfrac{49}{64}. ⇒ ( 1 + 113 7 ) ( 1 − 113 7 ) ( 1 + 113 8 ) ( 1 − 113 8 ) ⇒ ( 1 ) 2 − ( 113 7 ) 2 ( 1 ) 2 − ( 113 8 ) 2 ⇒ 1 − 113 49 1 − 113 64 ⇒ 113 113 − 49 113 113 − 64 ⇒ 113 64 113 49 ⇒ 64 49 .
Hence, (1 + sin θ)(1 - sin θ) (1 + cos θ)(1 - cos θ) = 49 64 \dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}} = \dfrac{49}{64} (1 + cos θ)(1 - cos θ) (1 + sin θ)(1 - sin θ) = 64 49
(ii) Given,
⇒ cot θ = 7 8 \dfrac{7}{8} 8 7
⇒ cot2 θ = ( 7 8 ) 2 = 49 64 \Big(\dfrac{7}{8}\Big)^2 = \dfrac{49}{64} ( 8 7 ) 2 = 64 49
Hence, cot2 θ = 49 64 \dfrac{49}{64} 64 49
If 3 cot A = 4, check whether 1 − tan 2 A 1 + tan 2 A \dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} 1 + tan 2 A 1 − tan 2 A 2 A - sin2 A or not.
Answer
Given,
⇒ 3 cot A = 4
⇒ cot A = 4 3 \dfrac{4}{3} 3 4
Let us draw a right angle triangle ABC.
We know that,
cot A = Side adjacent to ∠A Side opposite to ∠A \dfrac{\text{Side adjacent to ∠A}}{\text{Side opposite to ∠A}} Side opposite to ∠A Side adjacent to ∠A
Substituting values, we get :
AB BC = 4 3 \dfrac{\text{AB}}{\text{BC}} = \dfrac{4}{3} BC AB = 3 4
Let AB = 4k and BC = 3k.
In right angle triangle ABC,
⇒ AC2 = AB2 + BC2
⇒ AC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 25 k 2 \sqrt{25k^2} 25 k 2
We know that,
tan A = 1 cot A = 1 4 3 = 3 4 \dfrac{1}{\text{cot A}} = \dfrac{1}{\dfrac{4}{3}} = \dfrac{3}{4} cot A 1 = 3 4 1 = 4 3
Substituting value of tan A in 1 − tan 2 A 1 + tan 2 A \dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} 1 + tan 2 A 1 − tan 2 A
⇒ 1 − tan 2 A 1 + tan 2 A ⇒ 1 − ( 3 4 ) 2 1 + ( 3 4 ) 2 ⇒ 1 − 9 16 1 + 9 16 ⇒ 16 − 9 16 16 + 9 16 ⇒ 7 16 25 16 ⇒ 7 25 . \Rightarrow \dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} \\[1em] \Rightarrow \dfrac{1 - \Big(\dfrac{3}{4}\Big)^2}{1 + \Big(\dfrac{3}{4}\Big)^2} \\[1em] \Rightarrow \dfrac{1 - \dfrac{9}{16}}{1 + \dfrac{9}{16}} \\[1em] \Rightarrow \dfrac{\dfrac{16 - 9}{16}}{\dfrac{16 + 9}{16}} \\[1em] \Rightarrow \dfrac{\dfrac{7}{16}}{\dfrac{25}{16}} \\[1em] \Rightarrow \dfrac{7}{25}. ⇒ 1 + tan 2 A 1 − tan 2 A ⇒ 1 + ( 4 3 ) 2 1 − ( 4 3 ) 2 ⇒ 1 + 16 9 1 − 16 9 ⇒ 16 16 + 9 16 16 − 9 ⇒ 16 25 16 7 ⇒ 25 7 .
We know that,
cos A = Side adjacent to ∠A Hypotenuse = A B A C = 4 k 5 k = 4 5 \dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4k}{5k} = \dfrac{4}{5} Hypotenuse Side adjacent to ∠A = A C A B = 5 k 4 k = 5 4
sin A = Side opposite to ∠A Hypotenuse = B C A C = 3 k 5 k = 3 5 \dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3k}{5k} = \dfrac{3}{5} Hypotenuse Side opposite to ∠A = A C BC = 5 k 3 k = 5 3
Substituting value of cos A and sin A in cos2 A - sin2 A, we get :
⇒ ( 4 5 ) 2 − ( 3 5 ) 2 ⇒ 16 25 − 9 25 ⇒ 7 25 . \Rightarrow \Big(\dfrac{4}{5}\Big)^2 - \Big(\dfrac{3}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{16}{25} - \dfrac{9}{25} \\[1em] \Rightarrow \dfrac{7}{25}. ⇒ ( 5 4 ) 2 − ( 5 3 ) 2 ⇒ 25 16 − 25 9 ⇒ 25 7 .
Hence, proved that 1 − tan 2 A 1 + tan 2 A \dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} 1 + tan 2 A 1 − tan 2 A 2 A - sin2 A.
In triangle ABC, right-angled at B, if tan A = 1 3 \dfrac{1}{\sqrt{3}} 3 1
(i) sin A cos C + cos A sin C
(ii) cos A cos C - sin A sin C
Answer
Let us consider a right angle triangle ABC.
Given,
tan A = 1 3 \dfrac{1}{\sqrt{3}} 3 1
We know that,
tan A = Side opposite to ∠A Side adjacent to ∠A \dfrac{\text{Side opposite to ∠A}}{\text{Side adjacent to ∠A}} Side adjacent to ∠A Side opposite to ∠A
Substituting values, we get :
1 3 = B C A B \dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB} 3 1 = A B BC
Let AB = 3 \sqrt{3} 3
In △ABC,
By pythagoras theorem,
⇒ AC2 = AB2 + BC2
⇒ AC2 = ( 3 k ) 2 + k 2 (\sqrt{3}k)^2 + k^2 ( 3 k ) 2 + k 2
⇒ AC2 = 3k2 + k2
⇒ AC2 = 4k2
⇒ AC = 4 k 2 \sqrt{4k^2} 4 k 2
We know that,
sin A = Side opposite to ∠A Hypotenuse = B C A C = k 2 k = 1 2 \dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2} Hypotenuse Side opposite to ∠A = A C BC = 2 k k = 2 1
cos A = Side adjacent to ∠A Hypotenuse = A B A C = 3 k 2 k = 3 2 \dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2} Hypotenuse Side adjacent to ∠A = A C A B = 2 k 3 k = 2 3
sin C = Side opposite to ∠C Hypotenuse = A B A C = 3 k 2 k = 3 2 \dfrac{\text{Side opposite to ∠C}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2} Hypotenuse Side opposite to ∠C = A C A B = 2 k 3 k = 2 3
cos C = Side adjacent to ∠C Hypotenuse = B C A C = k 2 k = 1 2 \dfrac{\text{Side adjacent to ∠C}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2} Hypotenuse Side adjacent to ∠C = A C BC = 2 k k = 2 1
(i) Substituting values of sin A, cos C, sin C and cos A in sin A cos C + cos A sin C, we get :
⇒ 1 2 × 1 2 + 3 2 × 3 2 ⇒ 1 4 + 3 4 ⇒ 4 4 ⇒ 1. \Rightarrow \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{1}{4} + \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{4}{4} \\[1em] \Rightarrow 1. ⇒ 2 1 × 2 1 + 2 3 × 2 3 ⇒ 4 1 + 4 3 ⇒ 4 4 ⇒ 1.
Hence, sin A cos C + cos A sin C = 1.
(ii) Substituting values of cos A, cos C, sin A and sin C in cos A cos C - sin A sin C, we get :
⇒ 3 2 × 1 2 − 1 2 × 3 2 ⇒ 3 4 − 3 4 ⇒ 0. \Rightarrow \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2} - \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} \\[1em] \Rightarrow 0. ⇒ 2 3 × 2 1 − 2 1 × 2 3 ⇒ 4 3 − 4 3 ⇒ 0.
Hence, cos A cos C - sin A sin C = 0.
In △ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Answer
Given,
⇒ PR + QR = 25 cm
⇒ PR = (25 - QR) cm
In right angle triangle PQR,
By pythagoras theorem,
⇒ PR2 = PQ2 + QR2
⇒ (25 - QR)2 = 52 + QR2
⇒ 625 + QR2 - 50QR = 25 + QR2
⇒ 50QR = 625 - 25 + QR2 - QR2
⇒ 50QR = 600
⇒ QR = 600 50 \dfrac{600}{50} 50 600
⇒ PR = 25 - QR = 25 - 12 = 13
We know that,
sin P = Side opposite to ∠P Hypotenuse = Q R P R = 12 13 \dfrac{\text{Side opposite to ∠P}}{\text{Hypotenuse}} = \dfrac{QR}{PR} = \dfrac{12}{13} Hypotenuse Side opposite to ∠P = PR QR = 13 12
cos P = Side adjacent to ∠P Hypotenuse = P Q P R = 5 13 \dfrac{\text{Side adjacent to ∠P}}{\text{Hypotenuse}} = \dfrac{PQ}{PR} = \dfrac{5}{13} Hypotenuse Side adjacent to ∠P = PR PQ = 13 5
tan P = Side opposite to ∠P Side adjacent to ∠P = Q R P Q = 12 5 \dfrac{\text{Side opposite to ∠P}}{\text{Side adjacent to ∠P}} = \dfrac{QR}{PQ} = \dfrac{12}{5} Side adjacent to ∠P Side opposite to ∠P = PQ QR = 5 12
Hence, sin P = 12 13 , cos P = 5 13 and tan P = 12 5 \dfrac{12}{13}, \text{ cos P} = \dfrac{5}{13} \text{ and tan P} = \dfrac{12}{5} 13 12 , cos P = 13 5 and tan P = 5 12
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12 5 \dfrac{12}{5} 5 12
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4 3 \dfrac{4}{3} 3 4
Answer
(i) We know that,
tan 45° = 1.
∴ tan A is not always less than 1.
Hence, above statement is false.
(ii) We know that,
sec A ≥ 1
Since, 12 5 \dfrac{12}{5} 5 12
∴ sec A = 12 5 \dfrac{12}{5} 5 12
Hence, above statement is true.
(iii) We know that,
cosine A is also referred as cos A.
Hence, above statement is false.
(iv) We know that,
cot A ≠ cot × A
Hence, above statement is false.
(v) We know that,
0 ≤ sin θ ≤ 1.
Since, 4 3 > 1 \dfrac{4}{3} \gt 1 3 4 > 1
∴ sin θ cannot be equal to 4 3 \dfrac{4}{3} 3 4
Hence, above statement is false.
Evaluate the following :
sin 60° cos 30° + sin 30° cos 60°
Answer
Substituting values, we get :
⇒ sin 60° cos 30° + sin 30° cos 60° = 3 2 × 3 2 + 1 2 × 1 2 = 3 4 + 1 4 = 4 4 = 1. \Rightarrow \text{sin 60° cos 30° + sin 30° cos 60°} = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1. ⇒ sin 60° cos 30° + sin 30° cos 60° = 2 3 × 2 3 + 2 1 × 2 1 = 4 3 + 4 1 = 4 4 = 1.
Hence, sin 60° cos 30° + sin 30° cos 60° = 1.
Evaluate the following :
2 tan2 45° + cos2 30° - sin2 60°
Answer
Substituting values, we get :
⇒ 2 tan 2 45 ° + cos 2 30 ° − sin 2 60 ° = 2 × ( 1 ) 2 + ( 3 2 ) 2 − ( 3 2 ) 2 = 2 × ( 1 ) + 3 4 − 3 4 = 2. \Rightarrow \text{2 tan}^2 45° + \text{ cos}^2 30° - \text{ sin}^2 60° = 2 \times (1)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2 \\[1em] = 2 \times (1) + \dfrac{3}{4} - \dfrac{3}{4} \\[1em] = 2. ⇒ 2 tan 2 45° + cos 2 30° − sin 2 60° = 2 × ( 1 ) 2 + ( 2 3 ) 2 − ( 2 3 ) 2 = 2 × ( 1 ) + 4 3 − 4 3 = 2.
Hence, 2 tan2 45° + cos2 30° - sin2 60° = 2.
Evaluate the following :
cos 45° sec 30° + cosec 30° \dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}} sec 30° + cosec 30° cos 45°
Answer
Substituting values, we get :
⇒ cos 45° sec 30° + cosec 30° = 1 2 2 3 + 2 = 1 2 2 + 2 3 3 = 3 2 ( 2 + 2 3 ) = 3 2 2 ( 1 + 3 ) = 3 2 2 ( 1 + 3 ) × 1 − 3 1 − 3 = 3 ( 1 − 3 ) 2 2 ( 1 − 3 ) = 3 ( 1 − 3 ) 2 2 × − 2 = − 3 ( 1 − 3 ) 4 2 = 3 − 3 4 2 = 3 − 3 4 2 × 2 2 = 3 2 − 6 8 . \Rightarrow \dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}} = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}} + 2} \\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2 + 2\sqrt{3}}{\sqrt{3}}} \\[1em] = \dfrac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} \\[1em] = \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \\[1em] = \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \times \dfrac{1 - \sqrt{3}}{1 - \sqrt{3}} \\[1em] = \dfrac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2}(1 - 3)} \\[1em] = \dfrac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2} \times -2} \\[1em] = \dfrac{-\sqrt{3}(1 - \sqrt{3})}{4\sqrt{2}} \\[1em] = \dfrac{3 - \sqrt{3}}{4\sqrt{2}} \\[1em] = \dfrac{3 - \sqrt{3}}{4\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\[1em] = \dfrac{3\sqrt{2} - \sqrt{6}}{8}. ⇒ sec 30° + cosec 30° cos 45° = 3 2 + 2 2 1 = 3 2 + 2 3 2 1 = 2 ( 2 + 2 3 ) 3 = 2 2 ( 1 + 3 ) 3 = 2 2 ( 1 + 3 ) 3 × 1 − 3 1 − 3 = 2 2 ( 1 − 3 ) 3 ( 1 − 3 ) = 2 2 × − 2 3 ( 1 − 3 ) = 4 2 − 3 ( 1 − 3 ) = 4 2 3 − 3 = 4 2 3 − 3 × 2 2 = 8 3 2 − 6 .
Hence, cos 45° sec 30° + cosec 30° = 3 2 − 6 8 . \dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}} = \dfrac{3\sqrt{2} - \sqrt{6}}{8}. sec 30° + cosec 30° cos 45° = 8 3 2 − 6 .
Evaluate the following :
sin 30° + tan 45° - cosec 60° sec 30° + cos 60° + cot 45° \dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}} sec 30° + cos 60° + cot 45° sin 30° + tan 45° - cosec 60°
Answer
Substituting values, we get :
⇒ sin 30° + tan 45° - cosec 60° sec 30° + cos 60° + cot 45° = 1 2 + 1 − 2 3 2 3 + 1 2 + 1 = 3 2 − 2 3 3 2 + 2 3 = 3 2 − 2 3 3 2 + 2 3 × 3 2 − 2 3 3 2 − 2 3 = ( 3 2 − 2 3 ) 2 ( 3 2 ) 2 − ( 2 3 ) 2 = ( 3 2 ) 2 + ( 2 3 ) 2 − 2 × 3 2 × 2 3 ( 3 2 ) 2 − ( 2 3 ) 2 = 9 4 + 4 3 − 2 3 9 4 − 4 3 = 27 + 16 − 24 3 12 27 − 16 12 = 43 − 24 3 11 . \Rightarrow \dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}} = \dfrac{\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} + \dfrac{1}{2} + 1} \\[1em] = \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} + \dfrac{2}{\sqrt{3}}} \\[1em] = \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} + \dfrac{2}{\sqrt{3}}} \times \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}} \\[1em] = \dfrac{\Big(\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}\Big)^2}{\Big(\dfrac{3}{2}\Big)^2 - \Big(\dfrac{2}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{2}{\sqrt{3}}\Big)^2 - 2 \times \dfrac{3}{2} \times \dfrac{2}{\sqrt{3}}}{\Big(\dfrac{3}{2}\Big)^2 - \Big(\dfrac{2}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\dfrac{9}{4} + \dfrac{4}{3} - 2\sqrt{3}}{\dfrac{9}{4} - \dfrac{4}{3}} \\[1em] = \dfrac{\dfrac{27 + 16 - 24\sqrt{3}}{12}}{\dfrac{27 - 16}{12}} \\[1em] = \dfrac{43 - 24\sqrt{3}}{11}. ⇒ sec 30° + cos 60° + cot 45° sin 30° + tan 45° - cosec 60° = 3 2 + 2 1 + 1 2 1 + 1 − 3 2 = 2 3 + 3 2 2 3 − 3 2 = 2 3 + 3 2 2 3 − 3 2 × 2 3 − 3 2 2 3 − 3 2 = ( 2 3 ) 2 − ( 3 2 ) 2 ( 2 3 − 3 2 ) 2 = ( 2 3 ) 2 − ( 3 2 ) 2 ( 2 3 ) 2 + ( 3 2 ) 2 − 2 × 2 3 × 3 2 = 4 9 − 3 4 4 9 + 3 4 − 2 3 = 12 27 − 16 12 27 + 16 − 24 3 = 11 43 − 24 3 .
Hence, sin 30° + tan 45° - cosec 60° sec 30° + cos 60° + cot 45° = 43 − 24 3 11 . \dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}} = \dfrac{43 - 24\sqrt{3}}{11}. sec 30° + cos 60° + cot 45° sin 30° + tan 45° - cosec 60° = 11 43 − 24 3 .
Evaluate the following :
5 cos 2 60 ° + 4 sec 2 30 ° − tan 2 45 ° sin 2 30 ° + cos 2 30 ° \dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} sin 2 30° + cos 2 30° 5 cos 2 60° + 4 sec 2 30° − tan 2 45°
Answer
Solving,
⇒ 5 cos 2 60 ° + 4 sec 2 30 ° − tan 2 45 ° sin 2 30 ° + cos 2 30 ° ⇒ 5 × ( 1 2 ) 2 + 4 × ( 2 3 ) 2 − 1 2 ( 1 2 ) 2 + ( 3 2 ) 2 ⇒ 5 × 1 4 + 4 × 4 3 − 1 1 4 + 3 4 ⇒ 5 4 + 16 3 − 1 4 4 ⇒ 15 + 64 − 12 12 1 ⇒ 67 12 . \Rightarrow \dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} \\[1em] \Rightarrow \dfrac{5 \times \Big(\dfrac{1}{2}\Big)^2 + 4 \times \Big(\dfrac{2}{\sqrt{3}}\Big)^2 - 1^2}{\Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2} \\[1em] \Rightarrow \dfrac{5 \times \dfrac{1}{4} + 4 \times \dfrac{4}{3} - 1}{\dfrac{1}{4} + \dfrac{3}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{5}{4} + \dfrac{16}{3} - 1}{\dfrac{4}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{15 + 64 - 12}{12}}{1} \\[1em] \Rightarrow \dfrac{67}{12}. ⇒ sin 2 30° + cos 2 30° 5 cos 2 60° + 4 sec 2 30° − tan 2 45° ⇒ ( 2 1 ) 2 + ( 2 3 ) 2 5 × ( 2 1 ) 2 + 4 × ( 3 2 ) 2 − 1 2 ⇒ 4 1 + 4 3 5 × 4 1 + 4 × 3 4 − 1 ⇒ 4 4 4 5 + 3 16 − 1 ⇒ 1 12 15 + 64 − 12 ⇒ 12 67 .
Hence, 5 cos 2 60 ° + 4 sec 2 30 ° − tan 2 45 ° sin 2 30 ° + cos 2 30 ° = 67 12 \dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} = \dfrac{67}{12} sin 2 30° + cos 2 30° 5 cos 2 60° + 4 sec 2 30° − tan 2 45° = 12 67
Choose the correct option and justify your choice :
2 tan 30° 1 + tan 2 30 ° = \dfrac{\text{2 tan 30°}}{\text{1 + tan}^2 30°} = 1 + tan 2 30° 2 tan 30° =
sin 60°
cos 60°
tan 60°
sin 30°
Answer
Solving,
⇒ 2 tan 30° 1 + tan 2 30 ° = 2 × 1 3 1 + ( 1 3 ) 2 = 2 3 1 + 1 3 = 2 3 3 + 1 3 = 2 3 4 3 = 2 × 3 4 × 3 = 3 2 = sin 60° . \Rightarrow \dfrac{\text{2 tan 30°}}{\text{1 + tan}^2 30°} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{3 + 1}{3}} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}} \\[1em] = \dfrac{2 \times 3}{4 \times \sqrt{3}} \\[1em] = \dfrac{\sqrt{3}}{2} \\[1em] = \text{sin 60°}. ⇒ 1 + tan 2 30° 2 tan 30° = 1 + ( 3 1 ) 2 2 × 3 1 = 1 + 3 1 3 2 = 3 3 + 1 3 2 = 3 4 3 2 = 4 × 3 2 × 3 = 2 3 = sin 60° .
Hence, Option 1 is the correct option.
Choose the correct option and justify your choice :
1 - tan 2 45 ° 1 + tan 2 45 ° = \dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} = 1 + tan 2 45° 1 - tan 2 45° =
tan 90°
1
sin 45°
0
Answer
Solving,
⇒ 1 - tan 2 45 ° 1 + tan 2 45 ° = 1 − 1 1 + 1 = 0 2 = 0. \Rightarrow \dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} = \dfrac{1 - 1}{1 + 1} \\[1em] = \dfrac{0}{2} \\[1em] = 0. ⇒ 1 + tan 2 45° 1 - tan 2 45° = 1 + 1 1 − 1 = 2 0 = 0.
Hence, Option 4 is the correct option.
Choose the correct option and justify your choice :
sin 2A = 2 sin A is true when A =
0°
30°
45°
60°
Answer
Given,
Equation : sin 2A = 2 sin A
Substituting A = 0° in L.H.S. of above equation :
⇒ sin 2(0°) = sin 0° = 0.
Substituting A = 0° in R.H.S. of above equation :
⇒ 2 sin 0° = 2 sin 0° = 0.
Since, L.H.S. = R.H.S.
Hence, Option 1 is the correct option.
Choose the correct option and justify your choice :
2 tan 30° 1 - tan 2 30 ° = \dfrac{\text{2 tan 30°}}{\text{1 - tan}^2 30°} = 1 - tan 2 30° 2 tan 30° =
cos 60°
sin 60°
tan 60°
sin 30°
Answer
Solving,
⇒ 2 tan 30° 1 - tan 2 30 ° = 2 × 1 3 1 − ( 1 3 ) 2 = 2 3 1 − 1 3 = 2 3 2 3 = 2 × 3 2 × 3 = 3 = tan 60° . \Rightarrow \dfrac{\text{2 tan 30°}}{\text{1 - tan}^2 30°} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \\[1em] = \dfrac{2 \times 3}{2 \times \sqrt{3}} \\[1em] = \sqrt{3} \\[1em] = \text{tan 60°}. ⇒ 1 - tan 2 30° 2 tan 30° = 1 − ( 3 1 ) 2 2 × 3 1 = 1 − 3 1 3 2 = 3 2 3 2 = 2 × 3 2 × 3 = 3 = tan 60° .
Hence, Option 3 is the correct option.
If tan (A + B) = 3 \sqrt{3} 3 1 3 \dfrac{1}{\sqrt{3}} 3 1
Answer
Given,
⇒ tan (A + B) = 3 \sqrt{3} 3
⇒ tan (A + B) = tan 60°
⇒ A + B = 60° ..........(1)
Also,
⇒ tan (A - B) = 1 3 \dfrac{1}{\sqrt{3}} 3 1
⇒ tan (A - B) = tan 30°
⇒ A - B = 30° ..........(2)
Adding equation (1) and (2), we get :
⇒ A + B + A - B = 60° + 30°
⇒ 2A = 90°
⇒ A = 90 ° 2 \dfrac{90°}{2} 2 90°
Substituting value of A in equation (1), we get :
⇒ 45° + B = 60°
⇒ B = 60° - 45° = 15°.
Hence, A = 45° and B = 15°.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Answer
(i) Let A = 30° and B = 60°.
Substituting value of A and B in L.H.S. of the equation sin (A + B) = sin A + sin B, we get :
⇒ sin(A + B)
⇒ sin(30° + 60°)
⇒ sin 90°
⇒ 1.
Substituting value of A and B in R.H.S. of the equation sin (A + B) = sin A + sin B, we get :
⇒ sin A + sin B
⇒ sin 30° + sin 60°
⇒ 1 2 + 3 2 \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} 2 1 + 2 3
⇒ 1 + 3 2 \dfrac{1 + \sqrt{3}}{2} 2 1 + 3
Since, sin (A + B) ≠ sin A + sin B.
Hence, statement sin (A + B) = sin A + sin B is false.
(ii) Let θ be 0°, 30° and 45°.
sin 0° = 0, sin 30° = 1 2 \dfrac{1}{2} 2 1 1 2 \dfrac{1}{\sqrt{2}} 2 1
We see that,
sin θ increases as θ increases.
Hence, the statement the value of sin θ increases as θ increases is true.
(iii) Let θ be 0°, 30° and 45°.
cos 0° = 1, cos 30° = 3 2 \dfrac{\sqrt{3}}{2} 2 3 1 2 \dfrac{1}{\sqrt{2}} 2 1
We see that,
cos θ decreases as θ increases.
Hence, the statement the value of cos θ increases as θ increases is false.
(iv) Let θ be 0°, 30° and 45°.
sin 0° = 0, sin 30° = 1 2 \dfrac{1}{2} 2 1 1 2 \dfrac{1}{\sqrt{2}} 2 1
cos 0° = 1, cos 30° = 3 2 \dfrac{\sqrt{3}}{2} 2 3 1 2 \dfrac{1}{\sqrt{2}} 2 1
From above we see that,
sin θ = cos θ, only when θ = 45°.
Hence, the statement sin θ = cos θ for all values of θ is false.
(v) We know that,
cot A = cos A sin A \dfrac{\text{cos A}}{\text{sin A}} sin A cos A
For A = 0°, sin A = 0.
∴ cot A is not defined.
Hence, the statement cot A is not defined for A = 0° is true.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer
Converting sin A in terms of cot A,
⇒ sin A ⇒ sin 2 A ⇒ sin 2 A 1 ⇒ sin 2 A sin 2 A + cos 2 A ⇒ 1 sin 2 A sin 2 A + cos 2 A sin 2 A ⇒ 1 1 + cot 2 A . \Rightarrow \text{sin A} \\[1em] \Rightarrow \sqrt{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{1} \\[1em] \Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{\sqrt{\text{sin}^2 A + \text{ cos}^2 A}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{\dfrac{\text{sin}^2 A}{\text{sin}^2 A} + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{1 + \text{cot}^2 A}}. ⇒ sin A ⇒ sin 2 A ⇒ 1 sin 2 A ⇒ sin 2 A + cos 2 A sin 2 A ⇒ sin 2 A sin 2 A + sin 2 A cos 2 A 1 ⇒ 1 + cot 2 A 1 .
Converting sec A in terms of cot A,
We know that,
⇒ sec 2 A = 1 + tan 2 A ⇒ sec 2 A = 1 + 1 cot 2 A ⇒ sec 2 A = cot 2 A + 1 cot 2 A ⇒ sec A = cot 2 A + 1 cot 2 A ⇒ sec A = 1 + cot 2 A cot A . \Rightarrow \text{sec}^2 A = \text{1 + tan}^2 A \\[1em] \Rightarrow \text{sec}^2 A = 1 + \dfrac{1}{\text{cot}^2 A}\\[1em] \Rightarrow \text{sec}^2 A = \dfrac{\text{cot}^2 A + 1}{\text{cot}^2 A} \\[1em] \Rightarrow \text{sec A} = \sqrt{\dfrac{\text{cot}^2 A + 1}{\text{cot}^2 A}} \\[1em] \Rightarrow \text{sec A} = \dfrac{\sqrt{1 + \text{cot}^2 A}}{\text{cot A}}. ⇒ sec 2 A = 1 + tan 2 A ⇒ sec 2 A = 1 + cot 2 A 1 ⇒ sec 2 A = cot 2 A cot 2 A + 1 ⇒ sec A = cot 2 A cot 2 A + 1 ⇒ sec A = cot A 1 + cot 2 A .
We know that,
tan A = 1 cot A \dfrac{1}{\text{cot A}} cot A 1
Hence, sin A = 1 1 + cot 2 A , sec A = 1 + cot 2 A cot A and tan A = 1 cot A . \dfrac{1}{\sqrt{1 + \text{cot}^2 A}}, \text{sec A} = \dfrac{\sqrt{1 + \text{cot}^2 A}}{\text{cot A}} \text{ and tan A} = \dfrac{1}{\text{cot A}}. 1 + cot 2 A 1 , sec A = cot A 1 + cot 2 A and tan A = cot A 1 .
Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer
We know that,
⇒ cos A = 1 sec A \dfrac{1}{\text{sec A}} sec A 1
By formula,
⇒ sin2 A + cos2 A = 1
⇒ sin2 A = 1 - cos2 A
⇒ sin2 A = 1 - 1 sec 2 A \dfrac{1}{\text{sec}^2 A} sec 2 A 1
⇒ sin2 A = sec 2 A − 1 sec 2 A \dfrac{\text{sec}^2 A - 1}{\text{sec}^2 A} sec 2 A sec 2 A − 1
⇒ sin A = sec 2 A − 1 sec A \dfrac{\sqrt{\text{sec}^2 A - 1}}{\text{sec A}} sec A sec 2 A − 1
By formula,
⇒ sec2 A - tan2 A = 1
⇒ tan2 A = sec2 A - 1
⇒ tan A = sec 2 A − 1 \sqrt{\text{sec}^2 A - 1} sec 2 A − 1
By formula,
⇒ cot A = 1 tan A = 1 sec 2 A − 1 \dfrac{1}{\text{tan A}} = \dfrac{1}{\sqrt{\text{sec}^2 A - 1}} tan A 1 = sec 2 A − 1 1
By formula,
⇒ cosec A = 1 sin A = 1 sec 2 A − 1 sec A = sec A sec 2 A − 1 \dfrac{1}{\text{sin A}} = \dfrac{1}{\dfrac{\sqrt{\text{sec}^2 A - 1}}{\text{sec A}}} = \dfrac{\text{sec A}}{\sqrt{\text{sec}^2 A - 1}} sin A 1 = sec A sec 2 A − 1 1 = sec 2 A − 1 sec A
Hence,
sin A = sec 2 A − 1 sec A , cos A = 1 sec A , tan A = sec 2 A − 1 , cot A = 1 sec 2 A − 1 , cosec A = sec A sec 2 A − 1 \text{sin A} = \dfrac{\sqrt{\text{sec}^2 A - 1}}{\text{sec A}}, \text{cos A} = \dfrac{1}{\text{sec A}}, \text{tan A} = \sqrt{\text{sec}^2 A - 1}, \text{ cot A} = \dfrac{1}{\sqrt{\text{sec}^2 A - 1}}, \text{cosec A} = \dfrac{\text{sec A}}{\sqrt{\text{sec}^2 A - 1}} sin A = sec A sec 2 A − 1 , cos A = sec A 1 , tan A = sec 2 A − 1 , cot A = sec 2 A − 1 1 , cosec A = sec 2 A − 1 sec A
Choose the correct option. Justify your choice.
9 sec2 A - 9 tan2 A =
1
9
8
0
Answer
Solving,
⇒ 9 sec2 A - 9 tan2 A
⇒ 9 (sec2 A - tan2 A)
By formula,
sec2 A - tan2 = 1
⇒ 9 × 1 = 9.
Hence, Option 2 is the correct option.
Choose the correct option. Justify your choice.
(1 + tan θ + sec θ)(1 + cot θ - cosec θ) =
0
1
2
-1
Answer
Solving,
⇒ (1 + tan θ + sec θ)(1 + cot θ - cosec θ) ⇒ ( 1 + sin θ cos θ + 1 cos θ ) ( 1 + cos θ sin θ − 1 sin θ ) ⇒ ( sin θ + cos θ + 1 cos θ ) ( sin θ + cos θ - 1 sin θ ) ⇒ (sin θ + cos θ) 2 − 1 2 sin θ. cos θ ⇒ sin 2 θ + cos 2 θ + 2 sin θ. cos θ - 1 sin θ. cos θ ⇒ 1 + 2 sin θ.cos θ - 1 sin θ. cos θ ⇒ 2 sin θ. cos θ sin θ.cos θ ⇒ 2. \Rightarrow \text{(1 + tan θ + sec θ)(1 + cot θ - cosec θ)} \\[1em] \Rightarrow \Big(1 + \dfrac{\text{sin θ}}{\text{cos θ}} + \dfrac{1}{\text{cos θ}}\Big)\Big(1 + \dfrac{\text{cos θ}}{\text{sin θ}} - \dfrac{1}{\text{sin θ}}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin θ + cos θ + 1}}{\text{cos θ}}\Big)\Big(\dfrac{\text{sin θ + cos θ - 1}}{\text{sin θ}}\Big) \\[1em] \Rightarrow \dfrac{\text{(sin θ + cos θ)}^2 - 1^2}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ + \text{cos}^2 θ + \text{2 sin θ. cos θ - 1}}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{1 + \text{2 sin θ.cos θ - 1}}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{\text{2 sin θ. cos θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow 2. ⇒ (1 + tan θ + sec θ)(1 + cot θ - cosec θ) ⇒ ( 1 + cos θ sin θ + cos θ 1 ) ( 1 + sin θ cos θ − sin θ 1 ) ⇒ ( cos θ sin θ + cos θ + 1 ) ( sin θ sin θ + cos θ - 1 ) ⇒ sin θ. cos θ (sin θ + cos θ) 2 − 1 2 ⇒ sin θ. cos θ sin 2 θ + cos 2 θ + 2 sin θ. cos θ - 1 ⇒ sin θ. cos θ 1 + 2 sin θ.cos θ - 1 ⇒ sin θ.cos θ 2 sin θ. cos θ ⇒ 2.
Hence, Option 3 is the correct option.
Choose the correct option. Justify your choice.
(sec A + tan A) (1 – sin A) =
sec A
sin A
cosec A
cos A
Answer
Solving,
⇒ (sec A + tan A) (1 – sin A) ⇒ ( 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 \text{(sec A + tan A) (1 – sin A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)\text{(1 - sin A)} \\[1em] \Rightarrow \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big)\text{(1 - sin A)} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A}. ⇒ (sec A + tan A) (1 – sin 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 .
Hence, Option 4 is the correct option.
Choose the correct option. Justify your choice.
1 + tan 2 A 1 + cot 2 A = \dfrac{\text{1 + tan}^2 A}{\text{1 + cot}^2 A} = 1 + cot 2 A 1 + tan 2 A =
sec2 A
-1
cot2 A
tan2 A
Answer
Solving,
⇒ 1 + tan 2 A 1 + cot 2 A ⇒ sec 2 A cosec 2 A ⇒ 1 cos 2 A 1 sin 2 A ⇒ sin 2 A cos 2 A ⇒ tan 2 A . \Rightarrow \dfrac{\text{1 + tan}^2 A}{\text{1 + cot}^2 A} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A}{\text{cosec}^2 A} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos}^2 A}}{\dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{tan}^2 A. ⇒ 1 + cot 2 A 1 + tan 2 A ⇒ cosec 2 A sec 2 A ⇒ sin 2 A 1 cos 2 A 1 ⇒ cos 2 A sin 2 A ⇒ tan 2 A .
Hence, Option 4 is the correct option.
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
(cosec θ - cot θ)2 = 1 - cos θ 1 + cos θ \dfrac{\text{1 - cos θ}}{\text{1 + cos θ}} 1 + cos θ 1 - cos θ
Answer
Given,
Equation : (cosec θ - cot θ)2 = 1 - cos θ 1 + cos θ \dfrac{\text{1 - cos θ}}{\text{1 + cos θ}} 1 + cos θ 1 - cos θ
Solving L.H.S. of the above equation :
⇒ ( 1 sin θ − cos θ sin θ ) 2 ⇒ ( 1 − cos θ sin θ ) 2 ⇒ ( 1 - cos θ ) 2 sin 2 θ ⇒ ( 1 - cos θ ) 2 1 − cos 2 θ ⇒ ( 1 - cos θ ) 2 (1 - cos θ)(1 + cos θ) ⇒ 1 - cos θ 1 + cos θ . \Rightarrow \Big(\dfrac{1}{\text{sin θ}} - \dfrac{\text{cos θ}}{\text{sin θ}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \text{cos θ}}{\text{sin θ}}\Big)^2 \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{\text{sin}^2 θ} \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{1 - \text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{\text{(1 - cos θ)(1 + cos θ)}} \\[1em] \Rightarrow \dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}. ⇒ ( sin θ 1 − sin θ cos θ ) 2 ⇒ ( sin θ 1 − cos θ ) 2 ⇒ sin 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 .
Since, L.H.S. = R.H.S.
Hence, proved that (cosec θ - cot θ)2 = 1 - cos θ 1 + cos θ \dfrac{\text{1 - cos θ}}{\text{1 + cos θ}} 1 + cos θ 1 - cos θ
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
cos A 1 + sin A + 1 + sin A cos A \dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} 1 + sin A cos A + cos A 1 + sin A
Answer
Given,
Equation : cos A 1 + sin A + 1 + sin A cos A \dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} 1 + sin A cos A + cos A 1 + sin A
Solving L.H.S. of the equation :
⇒ cos A 1 + sin A + 1 + sin A cos A ⇒ cos 2 A + (1 + sin A) 2 cos A(1 + sin A) ⇒ cos 2 A + 1 + sin 2 A + 2 sin A cos A(1 + sin A) ⇒ sin 2 A + cos 2 A + 1 + 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{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + \text{(1 + sin A)}^2}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + 1 + \text{sin}^2 A + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 1 + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{1 + 1 + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2 + \text{2 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 \text{2 sec A}. ⇒ 1 + sin A cos A + cos A 1 + sin A ⇒ cos A(1 + sin A) cos 2 A + (1 + sin A) 2 ⇒ cos A(1 + sin A) cos 2 A + 1 + sin 2 A + 2 sin A ⇒ cos A(1 + sin A) sin 2 A + cos 2 A + 1 + 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 cos A 1 + sin A + 1 + sin A cos A \dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} 1 + sin A cos A + cos A 1 + sin A
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
tan θ 1 - cot θ + cot θ 1 - tan θ \dfrac{\text{tan θ}}{\text{1 - cot θ}} + \dfrac{\text{cot θ}}{\text{1 - tan θ}} 1 - cot θ tan θ + 1 - tan θ cot θ
Answer
Solving,
⇒ sin θ cos θ 1 − cos θ sin θ + cos θ sin θ 1 − sin θ cos θ ⇒ sin θ cos θ sin θ - cos θ sin θ + cos θ sin θ cos θ - sin θ cos θ ⇒ sin 2 θ cos θ(sin θ - cos θ) + cos 2 θ sin θ(cos θ - sin θ) ⇒ sin 2 θ cos θ(sin θ - cos θ) − cos 2 θ sin θ(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 θ - cos θ) ⇒ ( sin 2 θ + cos 2 θ + sin θ cos θ ) sin θ cos θ ⇒ 1 + sin θ cos θ sin θ cos θ ⇒ 1 sin θ cos θ + sin θ cos θ sin θ cos θ ⇒ sec θ cosec θ + 1. \Rightarrow \dfrac{\dfrac{\text{sin θ}}{\text{cos θ}}}{1 - \dfrac{\text{cos θ}}{\text{sin θ}}} + \dfrac{\dfrac{\text{cos θ}}{\text{sin θ}}}{1 - \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{\text{sin θ - cos θ}}{\text{sin θ}}} + \dfrac{\dfrac{\text{cos θ}}{\text{sin θ}}}{\dfrac{\text{cos θ - sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ}{\text{cos θ(sin θ - cos θ)}} + \dfrac{\text{cos}^2 θ}{\text{sin θ(cos θ - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ}{\text{cos θ(sin θ - cos θ)}} - \dfrac{\text{cos}^2 θ}{\text{sin θ(sin θ - cos θ)}} \\[1em] \Rightarrow \dfrac{\text{sin}^3 θ - \text{cos}^3 θ}{\text{sin θ cos θ (sin θ - cos θ)}} \\[1em] \Rightarrow \dfrac{\text{(sin θ - cos θ)}(\text{sin}^2 θ + \text{cos}^2 θ + \text{sin θ cos θ})}{\text{sin θ cos θ (sin θ - cos θ)}} \\[1em] \Rightarrow \dfrac{(\text{sin}^2 θ + \text{cos}^2 θ + \text{sin θ cos θ})}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{1 + \text{sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{1}{\text{sin θ cos θ}} + \dfrac{\text{sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \text{sec θ cosec θ} + 1. ⇒ 1 − sin θ cos θ cos θ sin θ + 1 − cos θ sin θ sin θ cos θ ⇒ sin θ sin θ - cos θ cos θ sin θ + cos θ cos θ - sin θ sin θ cos θ ⇒ cos θ(sin θ - cos θ) sin 2 θ + sin θ(cos θ - sin θ) cos 2 θ ⇒ cos θ(sin θ - cos θ) sin 2 θ − sin θ(sin θ - cos θ) cos 2 θ ⇒ sin θ cos θ (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 + sin θ cos θ ⇒ sin θ cos θ 1 + sin θ cos θ sin θ cos θ ⇒ sec θ cosec θ + 1.
Since, L.H.S. = R.H.S.
Hence, proved that tan θ 1 - cot θ + cot θ 1 - tan θ \dfrac{\text{tan θ}}{\text{1 - cot θ}} + \dfrac{\text{cot θ}}{\text{1 - tan θ}} 1 - cot θ tan θ + 1 - tan θ cot θ
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
1 + sec A sec A = sin 2 A 1 - cos A \dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}} sec A 1 + sec A = 1 - cos A sin 2 A
Answer
To prove:
1 + sec A sec A = sin 2 A 1 - cos A \dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}} sec A 1 + sec A = 1 - cos A sin 2 A
Solving L.H.S. of the equation :
⇒ 1 + 1 cos A 1 cos A ⇒ cos A + 1 cos A 1 cos 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 \text{1 + cos A}. ⇒ cos A 1 1 + cos A 1 ⇒ cos A 1 cos A cos A + 1 ⇒ 1 + cos A .
Solving R.H.S. of the equation :
⇒ sin 2 A 1 - cos A ⇒ 1 − cos 2 A 1 - cos A ⇒ (1 - cos A)(1 + cos A) 1 - cos A ⇒ 1 + cos A . \Rightarrow \dfrac{\text{sin}^2 A}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{1 - \text{cos}^2 A}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A)(1 + cos A)}}{\text{1 - cos A}} \\[1em] \Rightarrow \text{1 + cos A}. ⇒ 1 - cos A sin 2 A ⇒ 1 - cos A 1 − cos 2 A ⇒ 1 - cos A (1 - cos A)(1 + cos A) ⇒ 1 + cos A .
Since, L.H.S. = R.H.S. = 1 + cos A.
Hence, proved that 1 + sec A sec A = sin 2 A 1 - cos A \dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - 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 expressions are defined.
cos A - sin A + 1 cos A + sin A - 1 \dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} cos A + sin A - 1 cos A - sin A + 1 2 A = 1 + cot2 A.
Answer
To prove:
cos A - sin A + 1 cos A + sin A - 1 \dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} cos A + sin A - 1 cos A - sin A + 1
We know that,
⇒ cosec2 A = 1 + cot2 A
⇒ 1 = cosec2 A - cot2 A .....(1)
Dividing numerator and denominator of L.H.S. of the given equation with sin A, we get :
⇒ cos A - sin A + 1 cos A + sin A - 1 ⇒ cos A sin A − sin A sin A + 1 sin A cos A sin A + sin A sin A − 1 sin A ⇒ cot A - 1 + cosec A cot A + 1 - cosec A \Rightarrow \dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos A}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{sin A}} + \dfrac{1}{\text{sin A}}}{\dfrac{\text{cos A}}{\text{sin A}} + \dfrac{\text{sin A}}{\text{sin A}} - \dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{cot A - 1 + cosec A}}{\text{cot A + 1 - cosec A}} ⇒ cos A + sin A - 1 cos A - sin A + 1 ⇒ sin A cos A + sin A sin A − sin A 1 sin A cos A − sin A sin A + sin A 1 ⇒ cot A + 1 - cosec A cot A - 1 + cosec A
Substituting value of 1 from equation (1) in numerator of above equation, we get :
⇒ cot A + cosec A − (cosec 2 A − cot 2 A ) cot A + 1 - cosec A ⇒ cot A + cosec A − (cosec A − cot A ) (cosec A + cot A) cot A + 1 - cosec A ⇒ (cosec A + cot A)[1 - (cosec A - cot A)] cot A - cosec A + 1 ⇒ (cosec A + cot A)(cot A - cosec A + 1) cot A - cosec A + 1 ⇒ cosec A + cot A . \Rightarrow \dfrac{\text{cot A + cosec A} - \text{(cosec}^2 A - \text{cot}^2 A)}{\text{cot A + 1 - cosec A}} \\[1em] \Rightarrow \dfrac{\text{cot A + cosec A} - \text{(cosec} A - \text{cot} A)\text{(cosec A + cot A)}}{\text{cot A + 1 - cosec A}} \\[1em] \Rightarrow \dfrac{\text{(cosec A + cot A)[1 - (cosec A - cot A)]}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \dfrac{\text{(cosec A + cot A)(cot A - cosec A + 1)}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \text{cosec A + cot A}. ⇒ cot A + 1 - cosec A cot A + cosec A − (cosec 2 A − cot 2 A ) ⇒ cot A + 1 - cosec A cot A + cosec A − (cosec A − cot A ) (cosec A + cot A) ⇒ cot A - cosec A + 1 (cosec A + cot A)[1 - (cosec A - cot A)] ⇒ cot A - cosec A + 1 (cosec A + cot A)(cot A - cosec A + 1) ⇒ cosec A + cot A .
Since, L.H.S. = R.H.S.
Hence, proved that cos A - sin A + 1 cos A + sin A - 1 \dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} cos A + sin A - 1 cos A - sin A + 1
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
1 + sin A 1 - sin A \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} 1 - sin A 1 + sin A
Answer
To prove:
1 + sin A 1 - sin A \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} 1 - sin A 1 + sin A
Solving L.H.S. of the above equation, we get :
⇒ 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}}{\text{1 - sin A}}} \times \sqrt{\dfrac{\text{1 + sin A}}{\text{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
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
sin θ - 2 sin 3 θ 2 cos 3 θ − cos θ \dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}} 2 cos 3 θ − cos θ sin θ - 2 sin 3 θ
Answer
To prove:
sin θ - 2 sin 3 θ 2 cos 3 θ − cos θ \dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}} 2 cos 3 θ − cos θ sin θ - 2 sin 3 θ
Solving L.H.S. of the above equation :
⇒ sin θ(1 - 2 sin 2 θ ) cos θ (2 cos 2 θ − 1 ) ⇒ sin θ[1 - 2 (1 - cos 2 θ ) ] cos θ (2 cos 2 θ − 1 ) ⇒ sin θ[1 - 2 + 2cos 2 θ ) ] cos θ (2 cos 2 θ − 1 ) ⇒ tan θ (2 cos 2 θ − 1 ) 2 cos 2 θ − 1 ⇒ tan θ . \Rightarrow \dfrac{\text{sin θ(1 - 2 sin}^2 θ)}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{sin θ[1 - 2 (1 - cos}^2 θ)]}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{sin θ[1 - 2 + 2cos}^2 θ)]}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{tan θ}\text{ (2 cos}^2 θ - 1)}{\text{2 cos}^2 θ - 1} \\[1em] \Rightarrow \text{tan θ}. ⇒ cos θ (2 cos 2 θ − 1 ) sin θ(1 - 2 sin 2 θ ) ⇒ cos θ (2 cos 2 θ − 1 ) sin θ[1 - 2 (1 - cos 2 θ )] ⇒ cos θ (2 cos 2 θ − 1 ) sin θ[1 - 2 + 2cos 2 θ )] ⇒ 2 cos 2 θ − 1 tan θ (2 cos 2 θ − 1 ) ⇒ tan θ .
Since, L.H.S. = R.H.S.
Hence, proved that sin θ - 2 sin 3 θ 2 cos 3 θ − cos θ \dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}} 2 cos 3 θ − cos θ sin θ - 2 sin 3 θ
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Answer
To prove:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Solving L.H.S. of the above equation :
⇒ (sin A + cosec A)2 + (cos A + sec A)2
⇒ sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
⇒ sin2 A + cos2 A + cosec2 A + sec2 A + 2 sin A × 1 sin A + 2 cos A × 1 cos A \times \dfrac{1}{\text{sin A}} + \text{ 2 cos A} \times \dfrac{1}{\text{cos A}} × sin A 1 + 2 cos A × cos A 1
⇒ 1 + cosec2 A + sec2 A + 2 + 2
⇒ 5 + (1 + cot2 A) + (1 + tan2 A)
⇒ 7 + tan2 A + cot2 A.
Since, L.H.S. = R.H.S.
Hence, proved that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A.
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
(cosec A – sin A)(sec A – cos A) = 1 tan A + cot A \dfrac{1}{\text{tan A + cot A}} tan A + cot A 1
Answer
To prove:
(cosec A – sin A)(sec A – cos A) = 1 tan A + cot A \dfrac{1}{\text{tan A + cot A}} tan A + cot A 1
Solving L.H.S. of the given equation,
⇒ ( 1 sin A − sin A ) ( 1 cos A − cos A ) ⇒ ( 1 − sin 2 A sin A ) ( 1 − cos 2 A cos A ) ⇒ cos 2 A sin A × sin 2 A cos A ⇒ sin A cos A . \Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big)\Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big) \\[1em] \Rightarrow \Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big)\Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big) \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{sin A cos A}. ⇒ ( sin A 1 − sin A ) ( cos A 1 − cos A ) ⇒ ( sin A 1 − sin 2 A ) ( cos A 1 − cos 2 A ) ⇒ sin A cos 2 A × cos A sin 2 A ⇒ sin A cos A .
Solving R.H.S. of the given equation,
⇒ 1 sin A cos A + cos A sin A ⇒ 1 sin 2 A + cos 2 A sin A cos A ⇒ sin A cos A sin 2 A + cos 2 A ⇒ sin A cos A . \Rightarrow \dfrac{1}{\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}} \\[1em] \Rightarrow \dfrac{\text{sin A cos A}}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] \Rightarrow \text{sin A cos A}. ⇒ cos A sin A + sin A cos A 1 ⇒ sin A cos A sin 2 A + cos 2 A 1 ⇒ sin 2 A + cos 2 A sin A cos A ⇒ sin A cos A .
Since, L.H.S. = R.H.S. = sin A cos A
Hence, proved that (cosec A – sin A)(sec A – cos A) = 1 tan A + cot A \dfrac{1}{\text{tan A + cot A}} tan A + cot A 1
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
( 1 + tan 2 A 1 + cot 2 A ) = ( 1 − tan A 1 - cot A ) 2 \Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) = \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2 ( 1 + cot 2 A 1 + tan 2 A ) = ( 1 - cot A 1 − tan A ) 2 2 A
Answer
To prove:
( 1 + tan 2 A 1 + cot 2 A ) = ( 1 − tan A 1 - cot A ) 2 \Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) = \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2 ( 1 + cot 2 A 1 + tan 2 A ) = ( 1 - cot A 1 − tan A ) 2 2 A
Solving L.H.S. of the equation, we get :
⇒ ( 1 + tan 2 A 1 + cot 2 A ) ⇒ sec 2 A cosec 2 A ⇒ 1 cos 2 A 1 sin 2 A ⇒ sin 2 A cos 2 A ⇒ tan 2 A . \Rightarrow \Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) \\[1em] \Rightarrow \dfrac{\text{sec}^2 A}{\text{cosec}^2 A} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos}^2 A}}{\dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{tan}^2 A. ⇒ ( 1 + cot 2 A 1 + tan 2 A ) ⇒ cosec 2 A sec 2 A ⇒ sin 2 A 1 cos 2 A 1 ⇒ cos 2 A sin 2 A ⇒ tan 2 A .
Solving mid-portion, we get :
⇒ ( 1 − tan A 1 - cot A ) 2 ⇒ ( 1 − sin A cos A 1 − cos A sin A ) 2 ⇒ ( cos A - sin A cos A sin A - cos A sin A ) 2 ⇒ ( sin A(cos A - sin A) -cos A(cos A - sin A) ) 2 ⇒ ( -tan A ) 2 ⇒ tan 2 A . \Rightarrow \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2 \\[1em] \Rightarrow \Bigg(\dfrac{1 - \dfrac{\text{sin A}}{\text{cos A}}}{1 - \dfrac{\text{cos A}}{\text{sin A}}}\Bigg)^2 \\[1em] \Rightarrow \Bigg(\dfrac{\dfrac{\text{cos A - sin A}}{\text{cos A}}}{\dfrac{\text{sin A - cos A}}{\text{sin A}}}\Bigg)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin A(cos A - sin A)}}{\text{-cos A(cos A - sin A)}}\Big)^2 \\[1em] \Rightarrow (\text{-tan A})^2 \\[1em] \Rightarrow \text{tan}^2 A. ⇒ ( 1 - cot A 1 − tan A ) 2 ⇒ ( 1 − sin A cos A 1 − cos A sin A ) 2 ⇒ ( sin A sin A - cos A cos A cos A - sin A ) 2 ⇒ ( -cos A(cos A - sin A) sin A(cos A - sin A) ) 2 ⇒ ( -tan A ) 2 ⇒ tan 2 A .
Hence, proved that ( 1 + tan 2 A 1 + cot 2 A ) = ( 1 − tan A 1 - cot A ) 2 \Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) = \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2 ( 1 + cot 2 A 1 + tan 2 A ) = ( 1 - cot A 1 − tan A ) 2 2 A.
