Question

To prove that
$(\frac{1}{{\sec }^{2}x-{\cos }^{2}x}+\frac{1}{{\csc }^{2}x-{\sin }^{2}x}){\sin }^{2}x.{\cos }^{2}x$
$=\frac{1-{\sin }^{2}x{\cos }^{2}x}{2+\sin x{\cos }^{2}x}$

Answer 1

$[\frac{1}{({\sec }^{2}x-{\cos }^{2}x)}+\frac{1}{({\csc }^{2}x-{\sin }^{2}x)}]{\sin }^{2}x{\cos }^{2}x$

$=[\frac{{\cos }^{2}x}{1-{\cos }^{4}x}+\frac{{\sin }^{2}x}{1-{\sin }^{4}x}]{\sin }^{2}x{\cos }^{2}x$

$=\left[\frac{{\cos }^{2}x-{\cos }^{2}x{\sin }^{2}x-{\sin }^{2}x{\cos }^{4}x}{1+{\sin }^{4}x{\cos }^{4}x-{\cos }^{4}x-{\sin }^{4}x{\sin }^{2}x{\cos }^{2}x}\right]$

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

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