$\alpha$ to $\beta$

Since we are dealing with positive real numbers, we can consider using AM-GM inequality as follows:

$\begin{aligned} \frac {\cos^4 \alpha}{\cos^2 \beta} + \frac {\sin^4 \alpha}{\sin^2 \beta} & \ge 2\cdot \frac {\cos^2 \alpha \sin^2 \alpha}{\cos \beta \sin \beta} \end{aligned}$

Equality occurs when

$\begin{aligned} \frac {\cos^4 \alpha}{\cos^2 \beta} & = \frac {\sin^4 \alpha}{\sin^2 \beta} \\ \implies \tan^2 \beta & = \tan^4 \alpha \\ \tan \beta & = \tan^2 \alpha \\ \implies \tan \alpha & = \tan \beta = 1 \\ \alpha & = \beta = \frac \pi 4 \end{aligned}$

And

$\begin{aligned} \frac {\cos^4 \alpha}{\cos^2 \beta} + \frac {\sin^4 \alpha}{\sin^2 \beta} & = 2\cdot \frac {\cos^2 \alpha \sin^2 \alpha}{\cos \beta \sin \beta} = 2\cdot \frac {\frac 12 \cdot \frac 12}{\frac 1{\sqrt 2} \cdot \frac 1{\sqrt 2}} = 1 \end{aligned}$

Since 1 is the minumum value of the LHS and it is unique, the solution of $\alpha = \beta = \frac \pi 4$ is also unique. Therefore,

$\begin{aligned} \frac {\cos^4 \beta}{\cos^2 \alpha} + \frac {\sin^4 \beta}{\sin^2 \alpha} & = \frac {\cos^4 \alpha}{\cos^2 \beta} + \frac {\sin^4 \alpha}{\sin^2 \beta} = \boxed{1} \end{aligned}$

Let alpha and beta be $x,y$ for the sake of keyboard language changing problems xD.

Then, Let $a=\sin ^2x \implies 1-a=\cos ^2x$

Similarly, Let $b=\sin ^2y \implies 1-b=\cos ^2y$

Then, we have: $\frac{\left(1-a\right)^2}{1-b}+\frac{a^2}{b}=1$

$\implies \left(1-2a+a^2\right)b+a^2\left(1-b\right)=b-b^2$

$\implies b-2ab+a^2b+a^2-a^2b=b-b^2$

$\implies a^2 + b^2 -2ab = 0$

$\implies a-b=0$

$\implies a=b \implies \sin ^2x=\sin ^2y$

Clearly, $1-a=1-b \implies \cos ^2x=\cos ^2y$

$\implies \frac{\cos ^4y}{\cos ^2x}+\frac{\sin ^4y}{\sin ^2x} = \frac{\cos ^4x}{\cos ^2x}+\frac{\sin ^4x}{\sin ^2x} = \cos ^2x+\sin ^2x = 1$