How do I prove this trig identity that contains 2 variables?

Question

$ \frac{\sin A \cos B + \cos A \sin B } {\cos A \cos B - \sin A \sin B} = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

I've tried multiplying by the reciprocal denominator on the left side to see if I could begin to simplify it to tangent, but I'm still confused. Any help would be appreciated.

Answer 1

$LHS=\frac{\sin(A+B)}{\cos(A+B)}=\tan (A+B)=RHS $ by the composite angle formula for $\tan$.

Answer 2

LHS=$\frac{\sin(A+B)}{\cos(A+B)}=\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$=RHS

Compound angle formulae of $\sin, \cos$ and $\tan$ are used.

Answer 3

On the right hand side, put those tangents in terms of $\sin$ and $\cos$, and then multiply through to clear the denominators: $$\frac{\tan A+\tan B}{1-\tan A\tan B} = \frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{1-\frac{\sin A\sin B}{\cos A\cos B}} = \frac{\sin A\cos B\frac{\cos A}{\cos A}+\sin B\cos A\frac{\cos B}{\cos B}}{\cos A\cos B-\sin A\sin B\frac{\cos A\cos B}{\cos A\cos B}}=\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B-\sin A\sin B}$$ Done.

These are, of course, two ways of expressing $\tan(A+B)$.