Question

Prove: ${(\tan \theta +\sec \theta )}^2+{(\tan \theta -\sec \theta )}^2=\frac{2(1+{\sin }^2\theta )}{{\cos }^2\theta }$

Answer 1

$LHS={\tan }^2\theta +{\sec }^2\theta +2\tan \theta \sec \theta +{\tan }^2\theta +{\sec }^2\theta -2\tan \theta \sec \theta$

$=2({\tan }^2\theta +{\sec }^2\theta )$

$=2(\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{1}{{\cos }^2\theta })$

$=2\left(\frac{1+{\sin }^2\theta }{{\cos }^2\theta }\right)=RHS$