Question:
Prove that $\frac{\tan A+\tan B}{\tan A-\tan B}=\frac{\sin (A+B)}{\sin (A-B)}$.
Solution:
$\mathrm{LHS}=\frac{\tan A+\tan B}{\tan A-\tan B}$
$=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}$
$=\frac{\frac{\sin A \cos B+\cos A \sin B}{\cos A \cos B}}{\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B}}$
$=\frac{\sin A \cos B+\cos A \sin B}{\sin A \cos B-\cos A \sin B}$
$=\frac{\sin (A+B)}{\sin (A-B)}$
= RHS
Hemce proved.