Question
Prove the following identities:
$\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}=\frac{\sin^2\theta}{(1+\cos\theta)^2}$

Answer

$\text{LHS}=\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}$
$=\frac{\sec\theta-1}{\sec\theta+1}=\frac{\big(\frac{1}{\cos\theta}-1\big)}{\big(\frac{1}{\cos\theta}+1\big)}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
$=\frac{(1+\cos\theta)}{(1+\cos\theta)}\times\frac{(1+\cos\theta)}{(1+\cos\theta)}$
$=\frac{1-\cos^2\theta}{(1+\cos\theta)^2}$
$=\frac{\sin^2\theta}{(1+\cos\theta)^2}$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$

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