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

Answer

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

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