Question Number 53447 by rajeshghorai130@gmail.com last updated on 22/Jan/19
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
$${a}={sin}^{\mathrm{2}} {x} \\ $$$${b}={sin}^{\mathrm{2}} {y} \\ $$$$\frac{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }{\mathrm{1}−{b}}+\frac{{a}^{\mathrm{2}} }{{b}}=\mathrm{1} \\ $$$${b}−\mathrm{2}{ab}+{a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}={b}−{b}^{\mathrm{2}} \\ $$$${b}−\mathrm{2}{ab}+{a}^{\mathrm{2}} −{b}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}={b} \\ $$$${sin}^{\mathrm{2}} {x}={sin}^{\mathrm{2}} {y}\:\:\:\left[{so}\:{x}={y}\right] \\ $$$$ \\ $$$$\frac{{cos}^{\mathrm{4}} {y}}{{cos}^{\mathrm{2}} {x}}+\frac{{sin}^{\mathrm{4}} {y}}{{sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{{cos}^{\mathrm{4}} {x}}{{cos}^{\mathrm{2}} {x}}+\frac{{sin}^{\mathrm{4}} {x}}{{sin}^{\mathrm{2}} {x}} \\ $$$$={cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}\:{proved} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$