Question

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

Answer 1

$\frac{1}{{\sec }^2\alpha -{\cos }^2\alpha }=\frac{{\cos }^2\alpha }{1-{\cos }^4\alpha }(\because \sec \alpha =\frac{1}{\cos \alpha })$

$\frac{1}{{\csc }^2\alpha -{\sin }^2\alpha }=\frac{{\sin }^2\alpha }{1-{\sin }^4\alpha }(\because \csc \alpha =\frac{1}{\sin \alpha })$

$\frac{1}{{\sec }^2\alpha -{\cos }^2\alpha }+\frac{1}{{\csc }^2\alpha -{\sin }^2\alpha }=\frac{{\cos }^2\alpha }{1-{\cos }^4\alpha }+\frac{{\sin }^2\alpha }{1-{\sin }^4\alpha }$

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

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

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

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

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

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