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  $\frac{a^2}{b^2}=$

 $a{\sin }^{2}x+b{\cos }^{2}x=c$
      $\Rightarrow a\tan H2x+b=c{\sec }^{2}x$
      $\Rightarrow (a-c){\tan }^{2}x=c-b\Rightarrow {\tan }^{2}x=\frac{c-b}{a-c}$
     Similarly  ${\tan }^{2}y=\frac{d-a}{b-d}$
     But  $a\tan x=b\tan y\Rightarrow \frac{a^2}{b^2}=\frac{{\tan }^{2}y}{{\tan }^{2}x}$