Question:
Prove that
$\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)$
Solution:
$=\frac{\sin x+\sin y}{\sin x-\sin y}$
$=\frac{2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}}$
$=\tan \frac{x+y}{2} \cot \frac{x-y}{2}$
Using the formula,
$\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$
$\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$
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