I have two implicit equations, $F(x,y) = 0$, $G(x,y) = 0$, and I want to prove that on $\Omega \subseteq \mathbb{R}^{2}$ they describe the same plane curve. In other words, I need to show that $(x,y) \in \Omega$ satisfies $F(x,y) = 0$ if and only if it satisfies $G(x,y) = 0$.
In general, it might not be possible to express $x$ or $y$ from either of the equations. How would then one go about proving this?
In my problem, the equations are $$F(x,y) = \frac{1-\cos{(y-x)}}{\cos^2{(2x)}} - \frac{2\sin{(y)}\sin{(x)}}{\sin^2{(2x)}} = 0$$ $$G(x,y) = \frac{1-\cos{(y-x)}}{\cos^2{(2x)}} - (1 - \cos{(y+x)}) = 0$$ and I'm only interested in $$\Omega = (0, \frac{\pi}{4}) \times (0, \frac{\pi}{2}).$$
I'm grateful for any ideas. There might be something obvious that I'm missing, but I've spent the past several hours trying to think of something and I wasn't able to make progress.
Below is a Desmos link showing the two curves.
https://www.desmos.com/calculator/jvny2nrjqq
Thank you.
Hint:
$$\frac{1-\cos{(y-x)}}{\cos^2{(2x)}} - \frac{2\sin{(y)}\sin{(x)}}{\sin^2{(2x)}}\\ = \frac{(1-\cos{(y-x)})\sin^2(2x) -(\cos{(y-x)-\cos(x+y))}\cos^2(2x)}{\sin^2{(2x)}\cos^2{(2x)}}\\ = \frac{\sin^2(2x) -\cos{(y-x)+\cos(x+y)}\cos^2(2x)}{\sin^2{(2x)}\cos^2{(2x)}}\\ = \frac{1 -\cos(y-x)+(\cos(x+y)-1)\cos^2(2x)}{\sin^2{(2x)}\cos^2{(2x)}}\\ = \frac1{\sin^2(2x)}\left[\frac{1 -\cos(y-x)}{\cos^2(2x)}+\cos(x+y)-1\right] $$