Chapter 13 Trigonometric Identities
Class 10th R. S. Aggarwal Maths Solution
CBSE Class 10 Maths
R. S. Aggarwal Solution
Trigonometric Identities Exercise Ex. 13A
Solution 1
L.H.S = sin4θ - cos4θ
= (sin2θ)2 - (cos2θ)2
= (sin2θ - cos2θ) (sin2θ + cos2θ) … ∵ a2 - b2 = (a - b)(a + b)
= 1 (sin2θ - cos2θ) …. ∵ sin2θ + cos2θ = 1
= 1 - cos2θ - cos2θ …. ∵ sin2θ = 1 - cos2θ
= 1 - 2 cos2θ
= R.H.S
Hence proved.
Solution 2
Hence proved.
Solution 3
Hence proved.
Solution 4
Hence proved.
Solution 5
Solution 6
Hence proved.
Solution 7
L.H.S = sec4 θ - tan4 θ
= (sec2 θ)2 - (tan2 θ)2
= (sec2 θ - tan2 θ) (sec2 θ + tan2 θ)
= 1 × (sec2 θ + tan2 θ) …. ∵ sec2 θ - tan2 θ = 1
= 1 + tan2 θ + tan2 θ …. ∵ sec2 θ = 1 + tan2 θ
= 1 + 2 tan2 θ
= R.H.S
Hence proved.
Solution 8
Hence proved.
Solution 9
∵ L.H.S = R.H.S
Hence proved.
Solution 10
∵ L.H.S = R.H.S
⇒ sin2 θ tan θ + cos2 θ cot θ + 2 sin θ cos θ = tan θ + cot θ
Solution 11
Hence proved.
Solution 12
Hence proved.
Solution 13
∵ L.H.S = R.H.S
⇒ 
Solution 14
Hence proved.
Solution 15
Solution 16
∵ L.H.S = R.H.S
Solution 17
∵ L.H.S = R.H.S
Solution 18
Hence proved.
Solution 19
Hence proved.
Solution 20
Hence proved.
Solution 21
Hence proved.
Solution 22
From (1) and (2), we get
L.H.S = 
= 1 - sin θ cos θ + 1 + sin θ cos θ
= 2
= R.H.S
Hence proved.
Solution 23
Hence proved.
Solution 24
Hence proved.
Solution 25
Hence proved.
Solution 26
Hence proved.
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31(i)
Solution 31(ii)
Solution 31(iii)
Trigonometric Identities Exercise Ex. 13B
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Hence proved.
Solution 6
Solution 7
Solution 8
Now,
Solution 9
We have,
x = (sec A + sin A) and y = (sec A - sin A)
⇒ x + y = 2 sec A ⇒ sec A = 
⇒
cos A = 
…. (1)
And x - y = 2 sin A ⇒ sin A =
….. (2)
Solution 10
Solution 11
Hence proved.
Solution 12
Hence proved.
Solution 13
We know that,
sec2 θ - tan2 θ = 1
⇒ (sec θ + tan θ) (sec θ - tan θ) = 1 … a2 - b2 = (a - b)(a + b)
⇒ p (sec θ - tan θ) = 1 …. ∵ (sec θ + tan θ) = p
⇒ (sec θ - tan θ) = 
We have,
(sec θ + tan θ) = p …. (1) and (sec θ - tan θ) =
…. (2)
Adding (1) and (2), we get
2 sec θ = 
⇒ cos θ = 
…. (3)
Subtracting (2) from (1), we get
2 tan θ = 
⇒ tan θ 
Solution 14
Hence proved.
Solution 15
Hence proved.
Solution 16
Now,
Trigonometric Identities Exercise MCQ
Solution 1
Solution 2
Correct Option: (c)
Solution 3
Solution 4
Correct Option: (d)
Solution 5
Correct Option: (b)
sin θ - cos θ = 0 …. Given
⇒ (sin θ - cos θ)2 = sin2 θ + cos2 θ - 2 sin θ cos θ
⇒ 0 = 1 - 2 sin θ cos θ …. ∵ sin2 θ + cos2 θ = 1
⇒ 2 sin θ cos θ = 1
⇒ sin θ cos θ = 
⇒ sin2 θ cos2 θ = 
∴ sin4 θ + cos4 θ
= (sin2 θ + cos2 θ)2 - 2 sin2 θ cos2 θ
= 
Solution 6
Correct Option: (c)
cos 9α = sin α
⇒ sin(90˚- 9α) = sin α
∴ 90˚- 9α = α ⇒ 10α = 90˚ ⇒ α = 9˚
∴ tan 5α = tan 5(9˚) = tan 45˚ = 1
Solution 7
Correct Option: (d)
Solution 8
Correct option: (a)
Solution 9
Solution 10
Correct Option: (b)
Solution 11
Correct option: (a)
Solution 12
Solution 13
Solution 14
Correct Option: (a)
cosec2 θ = 1 + cot2 θ = 
∴ 
Solution 15
Correct Option: (c)
Solution 16
Correct Option: (a)
In a ∆ABC, ∠C = 90˚ ⇒ ∠A + ∠B = 90˚
∴ cos(A + B) = cos 90˚ = 0
Solution 17
Correct Option: (a)
We have, cos A + cos2 A = 1 … (1)
⇒ cos A = 1 - cos2 A = sin2 A ⇒ sin4 A = cos2 A
∴ (sin2 A + sin4 A) = cos A + cos2 A = 1 …from (1)
Solution 18
Solution 19
Solution 20
Correct option: (c)
We know that
cosec2 x - cot2 x = 1
Solution 21
