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

 

S.No.	Column-I	S.No.	Column-II
A)	$\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)$	p)	$\pm \frac{\sqrt{13}}{12}$
B)	$\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$	q)	$\frac{2}{3}$
C)	$\tan \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)$	r)	$\pm \frac{3}{\sqrt{13}}$
D)	$\tan \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$	s)	$\pm \sqrt{\frac{131}{13}}$

$\begin{array}{l}\cos \mathrm{\alpha }+\cos \mathrm{\beta }=1/2\\ \Rightarrow 2\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\frac{1}{2}\\ \Rightarrow \sin \mathrm{\alpha }+\sin \mathrm{\beta }=1/3\Rightarrow 2\sin \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\frac{1}{3}\end{array}$
Dividing Eq. (ii) by Eq. (i), we get $\tan \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)=\frac{2}{3}$
$\Rightarrow \cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)=\pm \frac{3}{\sqrt{13}}\cos (\mathrm{\alpha }-\mathrm{\beta })=-\frac{59}{72}$
Now, $\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\pm \frac{\sqrt{13}}{12}\text{ or }\tan \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\pm \sqrt{\frac{131}{13}}$