Question

Prove that $(\tan \theta +\cot \theta {)}^2={\sec }^2\theta +cose{c}^2\theta$

Answer 1

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

L.H.S $(\tan \theta +\cot \theta {)}^2$

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

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

$={\sec }^2\theta +cose{c}^2\theta -2+2$

$={\sec }^2\theta +cose{c}^2\theta$

$=$R.H.S.