Finding the value of $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$. 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 
$\cos^2\alpha \left(\frac{\cos^2 \alpha}{\cos^2 \beta}\right)+ \sin^2\alpha \left(\frac{\sin^2 \alpha}{\sin^2\beta}\right)$

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
$$
\frac{(1-s)^2}{1-s}+\frac{s^2}{s}=1-s+s=1.
$$
