Prove $\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}$ How would I solve the following double angle identity. 
$$
\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}
$$
So far my work has been.
$$
\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B+\sin A\sin B}
$$
But what would I do to continue. 

What i get is, how to solve the problem?? Is that correct then here u are:
$$\dfrac{\dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y + \sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y} +\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y} {\cos x\cos y}+ \dfrac{\sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\tan x+\tan y}{1+\tan x\tan y}$$
