# TRIGONOMETRIC IDENTITIES PROBLEMS WITH SOLUTIONS

Trigonometric Identities Problems with Solutions :

In this section, you will learn how to solve problems using trigonometric identities.

## Trigonometric Identities Problems with Solutions

Problem 1 :

(i) If (cos α/cos β) = m and (cos α/ sin β) = n then prove that (m2 + n2) cos2β  =  n2

Solution :

m  =  (cos α/cos β)

Taking squares on both sides

m2  =  (cos α/cos β)2

m2  =  (cosα/cosβ) ----------(1)

(cos α/ sin β) = n

n2  =  (cos α/sin β)2

n2  =  (cosα/sinβ) ----------(2)

(1) + (2)

m+ n2  =  (cosα/sinβ) + (cosα/cosβ)

=  (cosα cosβ + cosα sinβ) / sinβ cosβ

=  cosα (cosβ + sinβ) / sinβ cosβ

=  cosα (1) / sinβ cosβ

m+ n2   =  cosα / sinβ cosβ

(m+ n2⋅ cosβ  =  (cosα / sinβ cosβ)⋅ cosβ

=  (cosα / sinβ)

=  n2

(ii)  If cotθ + tan θ = x and sec θ −cos θ = y , then prove that (x2y)2/3 - (xy2)2/3  =  1

x = cotθ + tan θ

x = (cos θ/sin θ) + (sin θ/cos θ)

x = (cosθ + sinθ)/sin θ cos θ

x = 1/sin θ cos θ

x2 = 1/sin2 θ cosθ

y  =  sec θ −cos θ

y  =  (1/cos θ) −cos θ

y  =  (1 - cos2 θ)/cos θ

y  =  sin2 θ/cos θ

y2  =  sin4 θ/cosθ

x2y  =   (1/sin2 θ cosθ)  (sin2 θ/cos θ)

x2y  =   (1/cosθ)

(x2y)2/3  =   [(1/cosθ)]2/3

=  1/cos2θ

(x2y)2/3  =  sec2θ   ----(1)

(xy2)  =  (1/sin θ cos θ)(sin4 θ/cosθ)

=  (sin3 θ/cosθ)

(xy2)2/3  =  (sin2 θ/cosθ)

(xy2)2/3  = tan2θ ---(2)

(1) - (2)

(x2y)2/3 - (x2y)2/3  =  1

sec2θ - tan2θ  =  1

Hence proved.

Problem 2 :

(i) If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 −1) = 2p

Solution :

p = sin θ + cos θ

p2 = (sin θ + cos θ)2

=  sinθ + cosθ + 2 sin θ cos θ

p2  =  1 + 2 sin θ cos θ

p2 - 1  =  1 + 2 sin θ cos θ - 1

p- 1  =  sin θ cos θ

q = sec θ + cosec θ

q(p2 −1)  =  (sec θ + cosec θ)(sin θ cos θ)

=  (1/cos θ) + (1/sin θ) (sin θ cos θ)

=  [(sin θ cos θ)/cos θ] + [(sin θ cos θ)/sin θ]

=  sin θ + cos θ

=  2 (sin θ + cos θ)

=  2 p

(ii)  If sin θ (1+ sin2 θ) = cos2 θ, then prove that cos6 θ    −4 cos4 θ + 8 cos2 θ = 4

Solution :

sin θ (1+ sin2 θ) = cos2 θ

sin θ + sin3 θ = 1 - sin2 θ

sin θ + sin2 θ + sin3 θ = 1

sin θ + sin3 θ = 1 - sin2 θ

taking squares on both sides,

(sin θ + sin3 θ) =  (1 - sin2 θ)

sin2θ + 2sin4θ + sin6θ  =  1 - 2sin2θ + sin4θ

sin2θ + 2sin2θ + 2sin4θ - sin4θ + sin6θ  =  1

3sin2θ + sin4θ + sin6θ  =  1

3(1 - cos2θ) + (1 - cos2θ)2 (1 - cos2θ)3  =  1

3 - 3cos2θ + 1 + cos4θ - 2cos2θ + 1 - 3cos2θ + 3cos4θ - cos6θ  =

4 - cos6θ + 4cos4θ - 8cos2θ  =  0

cos6 θ − 4 cos4 θ + 8 cos2 θ = 4

Hence proved.

Problem 3 :

If cos θ / (1 + sin θ) = 1/a, then prove that (a2 - 1)/(a2 + 1) = sin θ

Solution :

cos θ / (1 + sin θ) = 1/a

a  =  (1 + sin θ) / cos θ

a2  =  [(1 + sin θ) / cos θ]2

a2  =  [(1 + sin θ)2 / cos2 θ]

a2  =  (1 + sin2θ + 2sin θ)/cos2 θ]

a2 - 1  =  (1 + sin2θ + 2sin θ)/cos2 θ] - 1

a- 1  =  (1 + sin2θ + 2sin θ - cos2 θ)/cos2 θ

a- 1  =  (2 sin2θ + 2sin θ)/cos2 θ

a- 1  = sinθ (sinθ + 1)/cos2θ

a- 1  = sinθ (sinθ + 1)/(1 - sin2θ)

a- 1  = sinθ /(1 - sinθ)  -----(1)

a2 + 1  =  (1 + sin2θ + 2sin θ)/cos2 θ] + 1

=  (1 + sin2θ + 2sin θ + cos2 θ)/cos2 θ

=  (1 + 1 + 2sin θ )/cos2 θ

=  2(1 + sin θ )/(1-sin2 θ)

=  2/(1-sin θ)  ---(2)

(1)/(2)

(a- 1) / (a+ 1)  =  [sinθ /(1 - sinθ)] / [2/(1-sin θ)]

=  sinθ

Hnce proved. After having gone through the stuff given above, we hope that the students would have understood how to solve problems using trigonometric Identities.

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