Question

Prove that : $\frac{(1+{\tan }^2\theta )\cot \theta }{cose{c}^2\theta }=\tan \theta$

Answer 1

$\frac{(1+{\tan }^2\theta )\cot \theta }{cose{c}^2\theta }$

$[\because {\sec }^2\theta -{\tan }^2\theta =1]$

$\frac{{\sec }^2\theta .\cot \theta }{cose{c}^2\theta }=\frac{\cot \theta }{cose{c}^2\theta /{\sec }^2\theta }$

$=\frac{\cot \theta }{{\cot }^2\theta }=\frac{1}{\cot \theta }$

$=\tan \theta$.