If $\cos \alpha +\cos \beta =m$and $\sin \alpha +\sin \beta =n,$ then $\sin (\alpha +\beta )=$

$m=\cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$

$n=\sin \alpha +\sin \beta =2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$

$\therefore \frac{n}{m}=\tan \left(\frac{\alpha +\beta }{2}\right)$
Now,

$\sin \left(\alpha +\beta \right)=\frac{2\tan \left({\displaystyle \frac{\alpha +\beta }{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}=\frac{{\displaystyle \frac{2n}{m}}}{1+\left({\displaystyle \frac{n^2}{m^2}}\right)}=\frac{2mn}{m^2+n^2}$