Question

Prove that $\frac{{\tan }^2\theta }{\sec \theta -1}=1+sec\theta$

Answer 1

$\frac{{\tan }^2\theta }{\sec \theta -1}\frac{{\sec }^2-1}{\sec \theta -1}\frac{(\sec \theta -1)(\sec \theta +1)}{\sec \theta -1}\sec \theta +1$

.Hence proved.