Question

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

Answer 1

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

Taking L.H.S.,

$\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}$

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

$=\frac{1}{\cos \theta .\sin \theta }$

Now, taking R.H.S.,

$\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$

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

$=\frac{1}{\sin \theta .\cos \theta }$

Therefore, L.H.S. $=$ R.H.S., hence proved