(b) We have $\sin A = n\sin B \Rightarrow \frac{n}{1} = \frac{{\sin A}}{{\sin B}}$

$ \Rightarrow \frac{{n – 1}}{{n + 1}} = \frac{{\sin A – \sin B}}{{\sin A + \sin B}} $

$= \frac{{2\cos \frac{{A + B}}{2}\sin \frac{{A – B}}{2}}}{{2\sin \frac{{A + B}}{2}\cos \frac{{A – B}}{2}}}$

$ = \tan \frac{{A – B}}{2}\cot \frac{{A + B}}{2}$

$ \Rightarrow \frac{{n – 1}}{{n + 1}}\tan \left( {\frac{{A + B}}{2}} \right) = \tan \frac{{A – B}}{2}$ .