Question

The value $\frac{1}{(1+{tan}^2\theta )}+\frac{1}{(1+{cot}^2\theta )}$ is

Answer 1

We have $1+{\tan }^2\theta ={\sec }^2\theta ,1+{\cot }^2\theta ={\csc }^2\theta$

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

$=\frac{1}{{\sec }^2\theta }+\frac{1}{{\csc }^2\theta }$

$={\cos }^2\theta +{\sin }^2\theta =1$