Answer
As left side transforms into right side, hence given identity-
$\frac{\sec\theta + 1}{\tan\theta} $ = $ \frac{\tan\theta}{\sec\theta - 1}$ is true.
Work Step by Step
Given identity is-
$\frac{\sec\theta + 1}{\tan\theta} $ = $ \frac{\tan\theta}{\sec\theta - 1}$
Taking L.S.
$\frac{\sec\theta + 1}{\tan\theta} $
= $\frac{\sec\theta + 1}{\tan\theta} . \frac{\sec\theta - 1}{\sec\theta - 1} $
({Multiplying the numerator and denominator by , $(\sec\theta - 1)$})
= $\frac{ \sec^{2} \theta - 1^{2}} {\tan\theta (\sec\theta - 1)}$
{Recall $(a+b)(a-b) $ = $a^{2} - b^{2}$ }
= $\frac{ \tan^{2} \theta} {\tan\theta (\sec\theta - 1)}$
( From second Pythagorean identity, $\sec^{2}\theta - 1$ = $\tan^{2}\theta$)
=$ \frac{\tan\theta}{\sec\theta - 1}$
= R.S.
As left side transforms into right side, hence given identity-
$\frac{\sec\theta + 1}{\tan\theta} $ = $ \frac{\tan\theta}{\sec\theta - 1}$ is true.
