Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
Question.

Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$

solution:

It is known that

$\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$\therefore$ L.H.S. $=\frac{\sin x-\sin y}{\cos x+\cos y}$

$=\frac{2 \cos \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)}$

$=\frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)}$

$=\tan \left(\frac{x-y}{2}\right)=$ R.H.S
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