<s>[INST] Proving a trig identity: $\frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B}$ 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.
 [/INST] 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
$$\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\tan (A+B)}{\tan (A-B)} \frac{1 - \tan A \tan B}{1 + \tan A \tan B} = \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B}$$
$$= \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos (A+B)}{\cos (A-B)} = \frac{\sin (A+B)}{\sin (A-B)}.$$
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