• # question_answer 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

Solution :

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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