Question

Prove $\sqrt{{\sec }^2\theta +{\csc }^2\theta }=\tan \theta +\cot \theta$

Answer 1

L.H.S$=\sqrt{{\sec }^2\theta +{\csc }^2\theta }$

$=\sqrt{1+{\tan }^2\theta +1+{\cot }^2\theta }$ since ${\sec }^2\theta =1+{\tan }^2\theta$ and ${\csc }^2\theta =1+{\cot }^2\theta$

$=\sqrt{{\tan }^2\theta +{\cot }^2\theta +2\tan \theta \cot \theta }$

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

$=\tan \theta +\cot \theta =$R.H.S

Hence proved