Question

Simplify:$\frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }$

Answer 1

Given: $\frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }$ = $\frac{{\sec }^2\theta }{{\csc }^2\theta }$ (By using trignometric identities)

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

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

$={\tan }^2\theta$