[tex]\dfrac{\sin(A+B)}{\sin(A-B)}=\dfrac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B}[/tex]
Divide through all terms by [tex]\cos A\cos B[/tex]. Then, for instance, the first term in the numerator becomes
[tex]\dfrac{\sin A\cos B}{\cos A\cos B}=\dfrac{\sin A}{\cos A}=\tan A[/tex]
All other terms reduce similarly, giving the final product
[tex]\dfrac{\tan A+\tan B}{\tan A-\tan B}[/tex]
