Question

Prove the following trigonometric identities. $\frac{1-sin\Theta}{1+sin\Theta}=(sec\Theta-tan\Theta)^{2}$

Answer 1

Answer with Step-by-step explanation:

Given : (1− sinθ)/(1+ sinθ) = (secθ − tanθ)²

L.H.S = (1− sinθ)/(1+ sinθ)

= [(1− sinθ)(1 - sinθ)] / ](1 + sinθ)(1− sinθ)]

[By rationalising ]

=  (1− sinθ)² /(1 - sin²θ)

=  (1− sinθ)² /(cos²θ)

[By using the identity, (1- sin²θ) = cos²θ]

= (1− sinθ / cosθ)²

= (1/cosθ − sinθ/cosθ)²

= (secθ − tanθ)²

[By using the identity, sec θ =1/cosθ & sinθ/cosθ = tanθ]

(1− sinθ)/(1+ sinθ) = (secθ − tanθ)²

L.H.S = R.H.S  

Hence Proved..

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Answer 2

Answer:

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