Question

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

Answer 1

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

$\frac{1}{\sec \theta -tan\theta }$$\times$$\left(\frac{\sec \theta +\tan \theta }{\sec \theta +\tan \theta }\right)$

$\frac{\sec \theta +tan\theta }{{\sec }^2\theta -ta{n}^2\theta }$

$=(\sec \theta +\tan \theta )$=rhs

LHS=RHS

Hence proved.