Question

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

Answer 1

We have,
LHS = $\frac{1}{\sec \theta -\tan \theta }$

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

$\Rightarrow$ LHS = $\frac{\sec \theta +\tan \theta }{{\sec }^2\theta -{\tan }^2\theta }=\frac{\sec \theta +\tan \theta }{1}$ $[\because {\sec }^2\theta -{\tan }^2\theta =1]$

$\Rightarrow$ LHS = $\sec \theta +\tan \theta =RHS$