Question

If $\frac{co{s}^{4}x}{co{s}^{2}y}+\frac{si{n}^{4}x}{si{n}^{2}y}=1$, Prove that $\frac{co{s}^{4}y}{co{s}^{2}x}+\frac{si{n}^{4}y}{si{n}^{2}x}=1$

Answer 1

A careful look at the question suggests that we have to prove x = y. We know that
$2co{s}^2\theta =1+cos2\theta ,$
and $2si{n}^2\theta =1-cos2\theta ,$
Hence changing to double angles in the give relation
$\frac{co{s}^{4}x}{co{s}^{2}y}+\frac{si{n}^{4}x}{si{n}^{2}y}=1$ we get
$\frac{1}{4}(1+cos2x{)}^2\frac{1}{2}(1-cos2y)+\frac{1}{4}(1-cos2x{)}^2\cdot \frac{1}{2}(1+cos2y)$
$\frac{1}{2}(1+cos2y)\cdot \frac{1}{2}(1-cos2y)$
$=2(1-co{s}^{2}2y)$
or $\{{(1+cos2x)}^2+{(1-cos2x)}^2\}-cos2y\{{(1+cos2x)}^2-{(1-cos2x)}^2\}$
$=2(1-co{s}^{2}2y)$
or $2(1+co{s}^{2}2x)-cos2y\left(4cos2x\right)=2-2co{s}^{2}2y$
or $co{s}^{2}2x+co{s}^{2}2y-2cos2xcos2y=0$
or ${(cos2x+cos2y)}^2=0$
$\therefore cos2x=cos2yory=n\pi \pm x$
$\therefore \frac{co{s}^{4}y}{co{s}^{2}x}+\frac{si{n}^{4}y}{si{n}^{2}x}=co{s}^{2}x+si{n}^{2}x=1$
(on putting y = x) .