# Prove that{1/(sec^2 a-cos^2 a) + 1/(cosec^2 a-sin^ a)}*cos^2 a sin^2 a=(1-cos^2 a sin^2 a)/(2+cos^2 a sin^2 a)

Aarushi Ahlawat
41 Points
6 years ago

First convert all the terms to sin and cos terms and then use identity to replace sin^2a+cos^a with 1. Here are the steps
$LHS=\left [ \frac{1}{sec^{2}(a)-cos^{2}(a)}+\frac{1}{cosec^{2}(a)-sin^{2}(a)} \right ]cos^{2}(a)sin^{2}(a)$

$=\left [ \frac{cos^{2}(a)}{1-cos^{4}(a)}+\frac{sin^{2}(a)}{1-sin^{4}(a)} \right ]cos^{2}(a)sin^{2}(a)$
$=\left [ \frac{cos^{2}(a)}{(1-cos^{2}(a))(1+cos^{2}(a))}+\frac{sin^{2}(a)}{(1-sin^{2}(a))(1+sin^{2}(a))} \right ]cos^{2}(a)sin^{2}(a)$
$=\left [ \frac{cos^{2}(a)}{sin^{2}(a)(1+cos^{2}(a))}+\frac{sin^{2}(a)}{cos^{2}(a)(1+sin^{2}(a))} \right ]cos^{2}(a)sin^{2}(a)$
$=\left [ \frac{cos^{2}(a)cos^{2}(a)(1+sin^{2}(a))+sin^{2}(a)sin^{2}(a)(1+cos^{2}(a))}{sin^{2}(a)(1+cos^{2}(a))cos^{2}(a)(1+sin^{2}(a))} \right ]cos^{2}(a)sin^{2}(a)$
$=\frac{cos^{4}(a)(1+sin^{2}(a))+sin^{4}(a)(1+cos^{2}(a))}{(1+cos^{2}(a))(1+sin^{2}(a))}$
$=\frac{cos^{4}(a)+cos^{4}(a)sin^{2}(a)+sin^{4}(a)+sin^{4}(a)cos^{2}(a))}{(1+cos^{2}(a)+sin^{2}(a)+cos^{2}(a)sin^{2}(a))}$
$=\frac{cos^{4}(a)+sin^{4}(a)+cos^{2}(a)sin^{2}(a)(sin^{2}(a)+cos^{2}(a))} {(1+1+cos^{2}(a)sin^{2}(a))}$
$=\frac{cos^{4}(a)+sin^{4}(a)+cos^{2}(a)sin^{2}(a)} {(2+cos^{2}(a)sin^{2}(a))}$
$=\frac{cos^{4}(a)+sin^{4}(a)+2cos^{2}(a)sin^{2}(a)-cos^{2}(a)sin^{2}(a)} {(2+cos^{2}(a)sin^{2}(a))}$
$=\frac{(cos^{2}(a)+sin^{2}(a))^{2}-cos^{2}(a)sin^{2}(a)} {(2+cos^{2}(a)sin^{2}(a))}$
$=\frac{1-cos^{2}(a)sin^{2}(a)} {2+cos^{2}(a)sin^{2}(a)}$
=RHS