If A = n B, then n - 1 n + 1 , A + B 2 = - Vidyadip

MCQ

If $\sin A = n\sin B,$ then $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $

A
$\sin \frac{{A - B}}{2}$
✓
$\tan \frac{{A - B}}{2}$
C
$\cot \frac{{A - B}}{2}$
D
None of these

✓

Answer

Correct option: B.

$\tan \frac{{A - B}}{2}$

b
(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}$ .

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