Question

Prove that:
${\sec }^2\theta +{\csc }^2\theta ={\sec }^2\theta \times {\csc }^2\theta$

Answer 1

LHS:

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

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

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

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

$={\sec }^2\theta {\csc }^2\theta$

:RHS

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

Hence Proved