Question

Evaluate:
$\frac{tan\theta }{sec\theta +1}=\frac{sec\theta -1}{tan\theta }$

Answer 1

Given,

$\frac{\tan \theta }{\sec \theta +1}=\frac{\sec \theta -1}{\tan \theta }$

upon cross multiplication, we get,

$(\tan \theta {)}^2=(\sec \theta +1\left)\right(\sec \theta -1)$

${\tan }^2\theta ={\sec }^2\theta -1$

${\tan }^2\theta +1={\sec }^2\theta$

${\sec }^2\theta ={\sec }^2\theta$

LHS = RHS

Hence proved