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Question-53447




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^2 x  b=sin^2 y  (((1−a)^2 )/(1−b))+(a^2 /b)=1  b−2ab+a^2 b+a^2 −a^2 b=b−b^2   b−2ab+a^2 −b+b^2 =0  (a−b)^2 =0  a=b  sin^2 x=sin^2 y   [so x=y]    ((cos^4 y)/(cos^2 x))+((sin^4 y)/(sin^2 x))  =((cos^4 x)/(cos^2 x))+((sin^4 x)/(sin^2 x))  =cos^2 x+sin^2 x  =1 proved
$${a}={sin}^{\mathrm{2}} {x} \\ $$$${b}={sin}^{\mathrm{2}} {y} \\ $$$$\frac{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }{\mathrm{1}−{b}}+\frac{{a}^{\mathrm{2}} }{{b}}=\mathrm{1} \\ $$$${b}−\mathrm{2}{ab}+{a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}={b}−{b}^{\mathrm{2}} \\ $$$${b}−\mathrm{2}{ab}+{a}^{\mathrm{2}} −{b}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}={b} \\ $$$${sin}^{\mathrm{2}} {x}={sin}^{\mathrm{2}} {y}\:\:\:\left[{so}\:{x}={y}\right] \\ $$$$ \\ $$$$\frac{{cos}^{\mathrm{4}} {y}}{{cos}^{\mathrm{2}} {x}}+\frac{{sin}^{\mathrm{4}} {y}}{{sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{{cos}^{\mathrm{4}} {x}}{{cos}^{\mathrm{2}} {x}}+\frac{{sin}^{\mathrm{4}} {x}}{{sin}^{\mathrm{2}} {x}} \\ $$$$={cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}\:{proved} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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