If $\sin A = n\sin B,$ then $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
If $\sin A = n\sin B,$ then $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
- A
$\sin \frac{{A - B}}{2}$
- B
$\tan \frac{{A - B}}{2}$
- C
$\cot \frac{{A - B}}{2}$
- D
None of these
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