If

Given:

$\sin A+\sin B=\frac{1}{4}$     ....(i)

$\cos A+\cos B=\frac{1}{2}$    ....(ii)

Dividing (i) by (ii):

$\Rightarrow \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{\frac{1}{4}}{\frac{1}{2}}$

$\Rightarrow \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{1}{2}$

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

$\Rightarrow \frac{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{1}{2}$

$\Rightarrow \tan \left(\frac{A+B}{2}\right)=\frac{1}{2}$

Hence proved.