Prove that \(\displaystyle \frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}\). Can someone provide me some hints? I tried to manipulate the right-hand expression but got back to square one.

Start on the right hand side by writing \(\displaystyle \frac 1{\tan (x - y)}\) as \(\displaystyle \frac {\cos (x - y)}{\sin (x - y)}\), then apply the addition formula for each. Where can you get with that?

Start on the right hand side by writing \(\displaystyle \frac 1{\tan (x - y)}\) as \(\displaystyle \frac {\cos (x - y)}{\sin (x - y)}\), then apply the addition formula for each. Where can you get with that?