Question

Prove:

$se{c}^2\theta +cose{c}^2\theta =se{c}^2\theta \times cose{c}^2\theta$

Answer 1

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

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

We know that

$\sec \theta =\frac{1}{\cos \theta },\csc \theta =\frac{1}{\sin \theta }$

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

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

$\left({\sec }^2\theta \right)+\left({\csc }^2\theta \right)=\frac{1}{{\sin }^2\theta {\cos }^2\theta }[\because {\sin }^2\theta +{\cos }^2\theta =1]$

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

$({\sec }^2\theta +{\csc }^2\theta )={\sec }^2\theta .{\csc }^2\theta [\because {\sec }^2\theta =\frac{1}{{\cos }^2\theta }{\csc }^2\theta =\frac{1}{{\sin }^2\theta }]$