If $\sin A+\sin B=\alpha$ and $\cos A+\cos B=\beta$, then write the value of $\tan \left(\frac{A+B}{2}\right)$
Given:
sin A + sin B = α .....(i)
cos A + cos B = β .....(ii)
Dividing (i) by (ii):
$\Rightarrow \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{\alpha}{\beta}$
$\Rightarrow \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{\alpha}{\beta}$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right.$ and $\left.\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$\Rightarrow \frac{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{\alpha}{\beta}$
$\Rightarrow \tan \left(\frac{A+B}{2}\right)=\frac{\alpha}{\beta}$
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