[Math] Prove $\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}$

Question

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.

Answer 1

Now divide by $\cos A \cos B$ and you are there