Question

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

• A. $tan\left(\frac{A-B}{2}\right)$
• B. $cot\left(\frac{A-B}{2}\right)$
• C. $sin\left(\frac{A-B}{2}\right)$
• D. $cos\left(\frac{A-B}{2}\right)$

Answer 1

The correct option is A

$tan\left(\frac{A-B}{2}\right)$

We have
$\sin A=n\sin B\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B}$

$\Rightarrow$ $\frac{n-1}{n+1}$ = $\frac{\sin A-\sin B}{\sin A+\sin B}$
$\Rightarrow$ $\frac{n-1}{n+1}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}$

$\Rightarrow \frac{n-1}{n+1}=\tan \frac{A-B}{2}\cot \frac{A+B}{2}$

$\Rightarrow \frac{n-1}{n+1}\tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{A-B}{2}\right).$