If $a{\sin }^{2}x+b{\cos }^{2}x=c,b{\sin }^{2}y+a{\cos }^{2}y=d$ and $a \tan x=b \tan y,$ then $\frac{a^2}{b^2}$ is equal to

We have,

$\begin{array}{l}a{\sin }^{2}x+b{\cos }^{2}x=c\\ \Rightarrow (b-a){\cos }^{2}x=c-a\\ \Rightarrow (b-a)=(c-a)\left(1+{\tan }^{2}x\right.)\end{array}$

Now, 

$\begin{array}{l}b{\sin }^{2}y+a{\cos }^{2}y=d\\ \Rightarrow (a-b){\cos }^{2}y=d-b\\ \Rightarrow (a-b)=(d-b)\left(1+{\tan }^{2}y\right)\\ \Rightarrow {\tan }^{2}y=\frac{a-d}{d-b}\\ \therefore {\tan }^{2}x=\frac{b-c}{c-a}\text{ and }{\tan }^{2}y=\frac{a-d}{d-b}\end{array}$

$\Rightarrow \frac{{\tan }^{2}x}{{\tan }^{2}y}=\frac{(b-c)(d-b)}{(c-a)(a-d)}$                           ... (i)

But $a\tan x=b\tan y\Rightarrow \frac{\tan x}{\tan y}=\frac{b}{a}$                      ... (ii)

From (i) and (ii), we get

$\frac{b^2}{a^2}=\frac{(b-c)(d-b)}{(c-a)(a-d)}\Rightarrow \frac{a^2}{b^2}=\frac{(c-a)(a-d)}{(b-c)(d-b)}$