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 $\frac{a^2}{b^2}=$

$a{\tan }^{2}x+b=c{\sec }^{2}x$

$a{\tan }^{2}x+b=c\left(1+{\tan }^{2}x\right)$

$\left(a-c\right){\tan }^{2}x=c-b$

${\tan }^{2}x=\frac{c-b}{a-c}$

$b{\sin }^{2}y+a{\cos }^{2}y=d$

$b{\tan }^{2}y+a=d{\sec }^{2}y$

$b{\tan }^{2}y+a=d\left(1+{\tan }^{2}y\right)$

$\left(b-d\right){\tan }^{2}y=d-a$

${\tan }^{2}y=\frac{d-a}{b-d}$

Given

$a\tan x=b\tan y$

$a^2{\tan }^{2}x=b^2{\tan }^{2}y$

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

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