If $\cos \alpha +\cos \beta =1/2\text{and}\sin \alpha +\sin \beta =1/3$ , then 

	Column-I		Column-II
A	$\cos \left(\frac{\alpha +\beta }{2}\right)$	P	$\pm \frac{\sqrt{13}}{12}$
B	$\cos \left(\frac{\alpha -\beta }{2}\right)$	Q	$\frac{2}{3}$
C	$\tan \left(\frac{\alpha +\beta }{2}\right)$	R	$\pm \frac{3}{\sqrt{13}}$
D	$\tan \left(\frac{\alpha -\beta }{2}\right)$	S	$\pm \sqrt{\frac{131}{13}}$

 $\cos \alpha +\cos \beta =\frac{1}{2}$
 $\Rightarrow 2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{2}$
 $\Rightarrow \sin \alpha +\sin \beta =1/3\Rightarrow 2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{3}$
Dividing Eq. (ii) by Eq. (i), we get  $\tan \left(\frac{\alpha +\beta }{2}\right)=\frac{2}{3}$
 $\Rightarrow \cos \left(\frac{\alpha +\beta }{2}\right)=\pm \frac{3}{\sqrt{13}}$

$$\begin{gathered} {\left(\cos \alpha +\cos \beta \right)}^2+{\left(\sin \alpha +\sin \beta \right)}^2=\frac{1}{4}+\frac{1}{9} \\ \Rightarrow 2+2\cos \left(\alpha -\beta \right)=\frac{13}{36}\therefore 4{\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{13}{36} \end{gathered}$$

$\cos (\alpha -\beta )=-\frac{59}{72}$
Now,  $\cos \left(\frac{\alpha -\beta }{2}\right)=\pm \frac{\sqrt{13}}{12}$ or  $\tan \left(\frac{\alpha -\beta }{2}\right)=\pm \sqrt{\frac{131}{13}}$