Question

Prove that $(\frac{1}{se{c}^2\theta -co{s}^2\theta }+\frac{1}{cose{c}^2\theta -si{n}^2\theta })si{n}^2\theta {\cos }^2\theta =\frac{1-{\sin }^2\theta co{s}^2\theta }{2+si{n}^2\theta co{s}^2\theta }$

Answer 1

LHS$=(\frac{1}{se{c}^2\theta -co{s}^2\theta }+\frac{1}{cose{c}^2\theta -si{n}^2\theta })si{n}^2\theta {\cos }^2\theta$

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

$=\left(\frac{{\cos }^2\theta -{\sin }^4\theta {\cos }^2\theta +{\sin }^2\theta -{\sin }^2\theta {\cos }^4\theta }{(1-{\cos }^4\theta )(1-{\sin }^4\theta )}\right){\sin }^2\theta {\cos }^2\theta$

$=\left(\frac{1-{\sin }^2\theta {\cos }^2\theta ({\sin }^2\theta +{\cos }^2\theta )}{1-{\sin }^4\theta -{\cos }^4\theta +{\sin }^4\theta {\cos }^4\theta }\right){\sin }^2\theta {\cos }^2\theta$

$=\left(\frac{1-{\sin }^2\theta {\cos }^2\theta }{1-({\sin }^2\theta +{\cos }^2\theta {)}^2+2{\sin }^2\theta {\cos }^2\theta +{\sin }^4\theta {\cos }^4\theta }\right){\sin }^2\theta {\cos }^2\theta$

$=\left(\frac{1-{\sin }^2\theta {\cos }^2\theta }{2{\sin }^2\theta {\cos }^2\theta +{\sin }^4\theta {\cos }^4\theta }\right){\sin }^2\theta {\cos }^2\theta$

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

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

$=RHS$

Hence proved