Question

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

Answer 1

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

$=(\frac{1}{\frac{1}{co{s}^2\theta }-co{s}^2\theta }+\frac{1}{\frac{1}{si{n}^2\theta }-si{n}^2\theta })si{n}^2\theta co{s}^2\theta$

$=(\frac{1}{\frac{1-co{s}^4\theta }{co{s}^2\theta }}+\frac{1}{\frac{1-si{n}^4\theta }{si{n}^2\theta }})si{n}^2\theta co{s}^2\theta$

$=(\frac{co{s}^2\theta }{(1-co{s}^2\theta )(1+co{s}^2\theta )}+\frac{si{n}^2\theta }{(1-si{n}^2\theta )(1+si{n}^2\theta )})si{n}^2\theta co{s}^2\theta$

[Using $1-a^4=1-(a^2{)}^2=(1-a^2\left)\right(1+a^2\left)\right]$

$=(\frac{co{s}^2\theta }{si{n}^2\theta (1+co{s}^2\theta )}+\frac{si{n}^2\theta }{co{s}^2\theta (1+si{n}^2)})si{n}^2\theta co{s}^2\theta$

[Using $1-co{s}^2\theta =si{n}^2\theta and1-si{n}^2\theta =co{s}^2\theta ]$

$=\left(\frac{co{s}^4\theta (1+si{n}^2\theta )+si{n}^4\theta (1+co{s}^2\theta )}{si{n}^3\theta co{s}^2\theta (1+co{s}^2\theta )(1+si{n}^2\theta )}\right).si{n}^2\theta co{s}^2\theta$

$=\frac{co{s}^4\theta +si{n}^2\theta co{s}^4\theta +si{n}^4\theta +co{s}^2\theta si{n}^4\theta }{(1+co{s}^2\theta )(1+si{n}^2\theta )}$

$=\frac{(co{s}^2\theta {)}^2+(si{n}^2\theta {)}^2+2co{s}^2\theta si{n}^2\theta -2co{s}^2\theta si{n}^2\theta +si{n}^2\theta co{s}^4\theta +co{s}^2\theta si{n}^4\theta }{(1+co{s}^2\theta )(1+si{n}^2\theta )}$

(adding and subtracting $2co{s}^2\theta si{n}^2\theta )$

$=\frac{(co{s}^2\theta +si{n}^2\theta {)}^2-2co{s}^2\theta sin62\theta +si{n}^2\theta co{s}^2\theta (co{s}^2\theta +si{n}^2\theta )}{1+si{n}^2\theta +co{s}^2\theta +si{n}^2\theta co{s}^2\theta }$

$=\frac{1^2-2co{s}^2\theta si{n}^2\theta +sin62\theta co{s}^2\theta .1}{1+1+si{n}^2\theta co{s}^2\theta }=\frac{1-si{n}^2\theta co{s}^2\theta }{2+si{n}^2\theta co{s}^2\theta }=RHS$ Hence proved.