Question: Trigonometry

$\dfrac{\cos^4 \alpha}{\cos^2 \beta}+ \dfrac{\sin^4\alpha}{\sin^2\beta} = 1$ 

then the value of 

$\dfrac{\cos^4\beta}{\cos^2\alpha}+ \dfrac{\sin^4\beta}{\sin^2\alpha}$ is?

NOTE: can somebody help me 
$\cos^2\alpha \left(\frac{\cos^2 \alpha}{\cos^2 \beta}\right)+ \sin^2\alpha \left(\frac{\sin^2 \alpha}{\sin^2\beta}\right)$

Answer: 
Let $t=\sin^2(\alpha)$ and $s=\sin^2(\beta)$. Then, multiplying both sides of the given identity by $s(1-s)$ gives:
$$
(1-t)^2s+t^2(1-s)=s(1-s).
$$
Bringing the RHS over to the LHS simplifies to $(s-t)^2=0$ so $s=t$. Now, the expression you want to evaluate is just
$$
\frac{(1-s)^2}{1-s}+\frac{s^2}{s}=1-s+s=1.
$$
$\cos^2 \alpha \left(\frac{\cos^2\alpha}{\cos^2 \beta}\right) + \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\sin^2 \beta}\right)=1$.

It can be noticed that $\frac{\cos^2\alpha}{\cos^2 \beta}$ and $\frac{\sin^2 \alpha}{\sin^2 \beta}$ must be $1$ (Why?). 

This tells us that $\cos^2 \alpha = \cos^2 \beta$ and $\sin^2 \alpha = \sin^2\beta$.

Plug those two results in the required expression. It will then become a well-known expression with a well-known value.