Question

$(\frac{1}{{\sec }^2\alpha -{\cos }^2\alpha }+\frac{1}{\csc {}^2\alpha -{\sin }^2\alpha })\times {\sin }^2\alpha {\cos }^2\alpha =\frac{1-{\sin }^2\alpha {\cos }^2\alpha }{2+{\sin }^2\alpha {\cos }^2\alpha }$

Answer 1

$[\frac{1}{{\sec }^2\alpha -{\cos }^2\alpha }+\frac{1}{{\csc }^2\alpha -{\sin }^2\alpha }].{\sin }^2\alpha {\cos }^2\alpha$

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

$[\frac{{\cos }^2\alpha }{1-{\cos }^4\alpha }+\frac{{\sin }^2\alpha }{{\cos }^2\alpha (1+{\sin }^2\alpha )}]{\sin }^2\alpha {\cos }^2\alpha$

$=\frac{{\cos }^4\alpha }{1+{\cos }^2\alpha }+\frac{{\sin }^2\alpha }{1+{\sin }^2\alpha }=\frac{{\cos }^4\alpha (1+{\sin }^2\alpha )+{\sin }^4\alpha (1+{\cos }^2\alpha )}{2+{\sin }^2\alpha {\cos }^2\alpha }$

$=\frac{{\cos }^4\alpha +{\sin }^4\alpha +{\sin }^2\alpha {\cos }^2\alpha ({\cos }^2\alpha +{\sin }^2\alpha )}{2+{\sin }^2\alpha {\cos }^2\alpha }$

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