Proving a trig identity: $\frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B}$

We start with $\frac{\tan A + \tan B}{\tan A - \tan B}$:

$$\frac{\tan A + \tan B}{\tan A - \tan B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}=\frac{\sin A\cos B +\sin B\cos A}{\sin A\cos B -\sin B\cos A}=\frac{\sin(A+B)}{\sin(A-B)}.$$

Using the tangent addition formulas, we get

$$\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$

and

$$\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

From this, we get

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

$$= \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos (A+B)}{\cos (A-B)} = \frac{\sin (A+B)}{\sin (A-B)}.$$

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