Question

Solve :- $\sqrt{cose{c}^2\theta +{\sec }^2\theta }=?$

Answer 1

$=\sqrt{\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }}$
$=\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }}$
$=\sqrt{\frac{1}{{\sin }^2\theta {\cos }^2\theta }}$
$=\sqrt{{\text{cosec}}^2\theta {\sec }^2\theta }$
$=|\text{cosec}\theta \sec \theta |$