Using trigonometric identities to prove following equation... I'm stuck on this problem:  


  If
  $$a\sin^2x + b\cos^2x = c$$ 
  $$b\sin^2y + a\cos^2y = d$$
  $$a\tan x=b\tan y$$
  then prove that
  $$\frac{a^2}{b^2} = \frac{(d-a)(c-a)}{(b-c)(b-d)}$$
$${{a^2}\over{b^2}}={{\tan^2y}\over{\tan^2x}}={{\sin^2y\cos^2x}\over{\cos^2y\sin^2x}}$$
also
$$\cos^2x={{c-a}\over{b-a}}$$
$$\sin^2x={{c-b}\over{a-b}}$$
$$\cos^2y={{d-b}\over{a-b}}$$
$$\sin^2y={{d-a}\over{b-a}}$$

Therefore the result can easily be derived.

$${{a^2}\over{b^2}}={{(d-a)(c-a)(a-b)(a-b)}\over{(b-a)(b-a)(c-b)(d-b)}}$$
I think it's wrong.

Try $a=b=c=d=x=y=1.$