Question

$\mathrm{If}\cos A+\cos B=\frac{1}{2}\mathrm{and}\sin A+\sin B=\frac{1}{4},\mathrm{prove}\mathrm{that}\tan \left(\frac{A+B}{2}\right)=\frac{1}{2}.$

Answer 1

Given:
sin A + sin B = $\frac{1}{4}$ .....(i)
cos A + cos B =$\frac{1}{2}$ .....(ii)

Dividing (i) by (ii):

$$\begin{gathered} \Rightarrow \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}=\frac{1}{2}\left[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\mathrm{and}\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ \Rightarrow \frac{\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}=\frac{1}{2} \\ \Rightarrow \tan \left(\frac{A+B}{2}\right)=\frac{1}{2} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$