Question

Prove the following identity
$\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\tan \theta +\cot \theta$

Answer 1

We need to prove $\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\tan \theta +\cot \theta$

L.H.S. $=\sqrt{{\sec }^2\theta +cose{c}^2\theta }$
$=\sqrt{1+{\tan }^2\theta +1+{\cot }^2\theta }$
$=\sqrt{{\tan }^2\theta +{\cot }^2\theta +2}$
$=\sqrt{{\tan }^2\theta +{\cot }^2\theta +2\tan \theta \cot \theta }$
$=\sqrt{{(\tan \theta +\cot \theta )}^2}$

$=$ R.H.S.