Prove that

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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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