# TRIGONOMETRIC IDENTITIES PROBLEMS WITH SOLUTIONS

Problem 1 :

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)

Square both sides

m2  =  (cos α/cos β)2

m2  =  (cos2α/cos2β) -----(1)

n  =  (cos α/ sin β)

Square both sides.

n2  =  (cos α/sin β)2

n2  =  (cos2α/sin2β) -----(2)

(1) + (2) :

m+ n2  =  (cos2α/sin2β) + (cos2α/cos2β)

m+ n=  (cos2αcos2β + cos2αsin2β) / sin2βcos2β

m+ n=  cos2α(cos2β + sin2β) / sin2βcos2β

m+ n=  cos2α(1) / sin2βcos2β

m+ n2   =  cos2α / sin2βcos2β

(m+ n2)cosβ  =  (cos2α / sin2βcos2β)cos2β

(m+ n2)cosβ  =  cos2α / sin2β

(m+ n2)cosβ  =  (cosα/sinβ)2

(m+ n2)cosβ  =  n2

Problem 2 :

If cotθ + tan θ = x and secθ - cosθ = y, then prove that

(x2y)2/3 - (xy2)2/3  =  1

Finding x2 :

x  =  cotθ + tanθ

x  =  (cosθ/sinθ) + (sinθ/cosθ)

x  =  (cos2θ + sin2θ)/sinθcosθ

x  =  1/sinθcosθ

x2  =  1/sin2θcos2θ

Finding y2 :

y  =  secθ  - cosθ

y  =  (1/cosθ) - cosθ

y  =  (1 - cos2θ)/cosθ

y  =  sin2θ/cosθ

y2  =  sin4θ/cos2θ

Finding  (x2y)2/3 :

x2y  =  (1/sin2θcos2θ)(sin2θ/cosθ)

x2y  =  (1/cos3θ)

(x2y)2/3  =  [(1/cos3θ)]2/3

(x2y)2/3  =  1/cos2θ

(x2y)2/3  =  sec2θ  -----(1)

Finding (xy2)2/3 :

xy2  =  (1/sinθ cosθ)(sin4θ/cos2θ)

xy2  =  (sin3θ/cos3θ)

(xy2)2/3  =  (sin2θ/cos2θ)

(xy2)2/3  = tan2θ ---(2)

(1) - (2) :

(x2y)2/3 - (x2y)2/3  =  1

sec2θ - tan2θ  =  1

Problem 3 :

If sinθ + cosθ = p and secθ + cosecθ = q, then prove that

q(p2 −1)  =  2p

Solution :

p  =  sinθ + cosθ

p2  =  (sinθ + cosθ)2

p2  =  sin2θ + cos2θ + 2sinθcosθ

p2  =  1 + 2sinθcosθ

p- 1  =  1 + 2sinθcosθ - 1

p- 1  =  2sinθcosθ

q(p2 −1)  =  (secθ + cosecθ)(2sinθcosθ)

q(p2 −1)  =  (1/cosθ) + (1/sinθ)(2sinθcosθ)

q(p2 −1)  =  [(2sinθcosθ)/cosθ] + [(2sinθcosθ)/sinθ]

q(p2 −1)  =  2sinθ + 2cosθ

q(p2 −1)  =  2(sinθ + cosθ)

q(p2 −1)  =  2p

Problem 4 :

If sinθ(1 + sin2θ)  =  cos2θ, then prove that

cos6θ - 4cos4θ + 8cos2θ  =  4

Solution :

sinθ(1 + sin2θ)  =  cos2θ

sinθ + sin3θ  =  1 - sin2θ

sinθ + sin2θ + sin3θ  =  1

sinθ + sin3θ  =  1 - sin2θ

Square both sides.

(sinθ + sin3θ) =  (1 - sin2θ)

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  =  1

4 - cos6θ + 4cos4θ - 8cos2θ  =  0

cos6θ − 4cos4θ + 8cos2θ  =  4

Problem 5 :

If cosθ/(1 + sinθ) = 1/a, then prove that

(a2 - 1)/(a2 + 1)  =  sinθ

Solution :

cosθ / (1 + sinθ)  =  1/a

a  =  (1 + sin θ)/cosθ

Finding (a2 - 1) :

a2  =  [(1 + sinθ)/cosθ]2

a2  =  [(1 + sinθ)2 / cos2θ]

a2  =  (1 + sin2θ + 2sinθ)/cos2θ]

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

a- 1  =  (1 + sin2θ + 2sinθ - cos2θ)/cos2θ

a- 1  =  (2sin2θ + 2sinθ)/cos2θ

a- 1  = 2sinθ(sinθ + 1)/cos2θ

a- 1  = 2sinθ(sinθ + 1)/(1 - sin2θ)

a- 1  =  2sinθ /(1 - sinθ) -----(1)

Finding (a2 + 1) :

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

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

a+ 1  =  (1 + 1 + 2sin θ )/cos2 θ

a+ 1  =  2(1 + sin θ )/(1-sin2 θ)

a+ 1  =  2/(1-sin θ) -----(2)

(1)/(2) :

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

(a- 1)/(a+ 1)  =  sinθ Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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