Question

Prove the following identity $\frac{{\tan }^2\theta }{{\tan }^2\theta -1}+\frac{cose{c}^2\theta }{{\sec }^2\theta -cose{c}^2\theta }=\frac{1}{{\sin }^2\theta -{\cos }^2\theta }$

Answer 1

$\frac{{\tan }^2\theta }{{\tan }^2\theta -1}+\frac{cose{c}^2\theta }{{\sec }^2\theta -cose{c}^2\theta }$

$=\frac{{\sin }^2\theta }{{\sin }^2\theta -{\cos }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta -{\cos }^2\theta }$

$=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta -{\cos }^2\theta }$

$=\frac{1}{{\sin }^2\theta -{\cos }^2\theta }$