Simplifying equation using trigonometric sum and differences identities Recently started learning trigonometric identities in school, having problems with this one. Tried solving using the sum and differences identities but keep having repeating everything. Thanks.
$\frac{-\cos(x)\sin(x)\pm \cos(y)\sin(y)}{\sin^2(y)-\cos^2(x)}=\tan(x\mp y)$

We have
$$LHS = \frac{-\cos x \sin x + \cos y \sin y}{\sin^2y-\cos^2x} = \dfrac{1}{2}\dfrac{\sin(2y)-\sin(2x)}{\sin^2y-\cos^2y} = \dfrac{\sin(y-x)\cos(y+x)}{\sin^2y-\cos^2x} $$

$\sin(-z) = -\sin(z)$ So $\sin(y-x) = -\sin(x-y)$

$$\begin{align}& LHS = \dfrac{\sin(x-y)\cos(y+x)}{\color{blue}{\cos^2x-\sin^2y}} = \dfrac{\sin(x-y)\cos(x+y)}{\color{blue}{\cos(x+y)\cos(x-y)}}\\
& \ = \dfrac{\sin(x-y)}{\cos(x-y)} =\tan(x-y) =RHS\end{align}$$

Use $y = -y$ to get the other one.

For the blue highlighted part (a short proof)

$\cos(x+y)\cos(x-y) = \dfrac{1}{2}\left[\cos(2x)+\cos(2y)\right] = \dfrac{1}{2}\left[(2\cos^2x -1) + (1-2\sin^2y)\right]\\ = \cos^2x-\sin^2y$

