MCQ
જો $\sin A = n\sin B,$ તો $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
- A$\sin \frac{{A - B}}{2}$
- ✓$\tan \frac{{A - B}}{2}$
- C$\cot \frac{{A - B}}{2}$
- Dએકપણ નહિ.
$ \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}$ .
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