Simplifying $\frac {\cos^4 x}{\cos^2 y}＋\frac {\sin^4 x}{\sin^2 y}＝1$ How do you simplify the following?
$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$
What I've tried:
$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$
$$\sin^2(x)+\cos^2(x)=1$$
$$\implies\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝\sin^2(x)+\cos^2(x)$$
$$\implies \frac {\cos^4(x)}{\cos^2(y)}－\cos^2(x)＝\sin^2(x)－\frac {\sin^4(x)}{\sin^2(y)}\\~\\\implies \frac {\cos^4(x)－\cos^2(x)\cos^2(y)}{\cos^2(y)}＝\frac {\sin^2(x)\sin^2(y)－\sin^4(x)}{\sin^2(y)}\\~\\\implies \frac {\cos^2(x)}{\cos^2(y)}\left(\cos^2(x)－\cos^2(y)\right)＝\frac {\sin^2(x)}{\sin^2(y)}\left(\sin^2(y)－\sin^2(x)\right)$$
 A: \begin{eqnarray}\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}&=&1\\
\cos^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-\sin^2x)^2\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x+\sin^4x)\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x)\sin^2y+\sin^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x)\sin^2y+\sin^4x&=&\sin^2y(1-\sin^2y)\\
\sin^4x&=&\sin^2y(1-\sin^2y)-(1-2\sin^2x)\sin^2y\\
\sin^4x&=&(1-\sin^2y-1+2\sin^2x)\sin^2y\\
\sin^4x&=&2\sin^2x\sin^2y-\sin^4y\\
\sin^4x+\sin^4y&=&2\sin^2x\sin^2y\\
\frac{\sin^2x}{\sin^2y}+\frac{\sin^2y}{\sin^2x}&=&2
\end{eqnarray}
Let $u=\frac{\sin^2x}{\sin^2y}$, then $u+\frac{1}{u}=2$. So $(u-1)^2=0$.
Thus $u=1$ which gives $$\sin^2x=\sin^2y$$
Note that the original expression is undefined when either of $\cos y$ or $\sin y$ is zero whereas the 'equivalent' equation is defined for all $x,y$. If the original equation is set to $a$ instead of $1$ and we allow $a\to1$ we will see why.
The following desmos graph compares the graph for $a=1.2$ (in red) compared to the graph of $\sin^2x=\sin^2y$ (in black).
Here is a link to a graph in motion where $a$ varies in value from 1 to 2.

Desmos dynamic graph
