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Question
Prove that:
Solution
Consider LHS:
\[ \frac{\sin A + \sin B}{\sin A - \sin B}\]
\[ = \frac{2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}{2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)} \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right), and \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos\left( \frac{A + B}{2} \right) \right\}\]
\[ = \frac{\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}{\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}\]
\[ = \tan \left( \frac{A + B}{2} \right) cot \left( \frac{A - B}{2} \right)\]
= RHS
Hence, LHS = RHS.
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