I'm learning about trig identities, and I'm struggling to prove that two expressions are equal:
$$
\frac{\sin(A + B)}{\sin(A – B)}=\frac{\tan A + \tan B}{\tan A – \tan B}
$$
How do you go about proving this? I know about compound angles – i.e. the sine, cosine and tangent of $(A \pm B)$, but don't know how to apply it in this situation.
Best Answer
We start with $\frac{\tan A + \tan B}{\tan A - \tan B}$:
$$\frac{\tan A + \tan B}{\tan A - \tan B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}=\frac{\sin A\cos B +\sin B\cos A}{\sin A\cos B -\sin B\cos A}=\frac{\sin(A+B)}{\sin(A-B)}.$$