If sin A = nsin B, then frac{{n - 1}}{{n + 1} | vedclass

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Question

(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 - \s

Vedclass · Accepted Answer

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

Vedclass · Answer

(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}} $

$= \frac{{2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}}}{{2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2}}}$ 

$ = 	an \frac{{A - B}}{2}\cot \frac{{A + B}}{2}$ 

$ \Rightarrow \frac{{n - 1}}{{n + 1}}	an \left( {\frac{{A + B}}{2}} ight) = 	an \frac{{A - B}}{2}$ .

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

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