Question

Solve $\frac{(1+{\tan }^{2}Q)\cot Q}{{\csc }^{2}Q}=\tan Q$

Answer 1

$\frac{\left(1+{\tan }^2\theta \right)\cot \theta }{\cos e{c}^2\theta }$

$=\frac{{\sec }^2\theta {\cot }^2\theta }{\cos e{c}^2\theta }$

$=\frac{1}{{\cos }^2\theta }\times \frac{\cos \theta }{\sin \theta }\times {\sin }^2\theta$

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

$=\tan \theta$ Proved