show this identity with trigometric

Question

I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.

Problem::

let $x,y\in (0,\dfrac{\pi}{2})$. show that $$\dfrac{\sin{(x+y)}\tan{x}-\cos{(x+y)}}{\sin{(x+y)}\tan{y}-\cos{(x+y)}}=\dfrac{\cos{(2x+y)}\cos{y}}{\cos{(x+2y)}\cos{x}}$$

This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks

Answer 1

Hint:

$$\begin{align} \sin(x+y)\tan x - \cos(x+y) &= \phantom{-}\frac{1}{\cos x}\left(\;\sin(x+y) \sin x - \cos(x+y)\cos x\;\right) \\[4pt] &= -\frac1{\cos x}\cos\left((x+y)+x\right) \\[4pt] &= -\frac1{\cos x}\cos\left(2x+y\right) \end{align}$$