Question

Prove that $\frac{(1+{\cot }^2\theta )\tan \theta }{{\sec }^2\theta }=\cot \theta$.

Answer 1

Identity : $cose{c}^2\theta -{\cot }^2\theta =1$

LHS $=\frac{(1+{\cot }^2\theta )\tan \theta }{{\sec }^2\theta }$

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

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

$=\cot \theta$

$=RHS$

Hence proved