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सिद्ध करें कि \(\frac{\sin A + \sin B}{\cos A + \cos B} = \tan \frac{A+B}{2}\)

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\(\frac{\sin A + \sin B}{\cos A + \cos B} = \tan \frac{A+B}{2}\)  

L.H.S :-

\(\frac{\sin A + \sin B}{\cos A + \cos B} \)  

\(=\frac{2 \sin(\frac{A+B}{2})\cos(\frac{A-B}{2})}{2 \cos(\frac{A+B}{2})\cos(\frac{A-B}{2})}\)      \(\left[ \because \sin A+ \sin B = 2 \sin (\frac{A+B}{2}) \cos(\frac{A-B}{2})\\ \cos A+ \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})\right] \)

\(=\frac{\sin}{\cos}(\frac{A+B}{2})\)  

\(=\tan(\frac{A+B}{2})\)

= R.H.S

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