$ \Rightarrow \,\,\frac{{m + 1}}{{m - 1}} = \frac{{\cos A + \cos B}}{{\cos A - \cos B}} $
$= \frac{{2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{B - A}}{2}} \right)}}{{2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{B - A}}{2}} \right)}}$
$ = \cot \,\left( {\frac{{A + B}}{2}} \right)\,\cot \,\left( {\frac{{B - A}}{2}} \right)$
Hence, $\cot \,\left( {\frac{{A + B}}{2}} \right) = \frac{{m + 1}}{{m - 1}}\tan \frac{{B - A}}{2}$.
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