Question

If $\sin A=n\sin B$, then $\left[\frac{(n-1)}{(n+1)}\right]\tan \frac{(A+B)}{2}=$

• A. $\sin \left(\frac{A-B}{2}\right)$
• B. $\tan \left(\frac{A-B}{2}\right)$
• C. $cot\left(\frac{A-B}{2}\right)$
• D. None of these

Answer 1

The correct option is B

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

Explanation for the correct option.

Given that, $\sin A=n\sin B$

It can be rewritten as: $\frac{\sin A}{\sin B}=\frac{n}{1}$

By componendo and dividendo rule we have:
$\frac{(\mathrm{sinA}+\mathrm{sinB})}{(\mathrm{sinA}-\mathrm{sinB})}=\frac{(\mathrm{n}+1)}{(\mathrm{n}-1)}$
Now we know that,$\sin x+\sin y=2\sin \frac{(x+y)}{2}\cos \frac{(x-y)}{2}\mathrm{and}\sin x-\sin y=2\sin \frac{(x-y)}{2}\cos \frac{(x+y)}{2}$

So we have
$\begin{array}{rcl}\frac{2\sin \frac{(A+B)}{2}\cos \frac{(A-B)}{2}}{2\sin {\displaystyle \frac{(A-B)}{2}}\cos {\displaystyle \frac{(A+B)}{2}}}& =& \frac{(n+1)}{(n-1)}\\ \frac{\tan \frac{(A+B)}{2}}{\tan {\displaystyle \frac{(A-B)}{2}}}& =& \frac{(n+1)}{n-1}\\ \frac{n-1}{n+1}\tan \left(\frac{A+B}{2}\right)& =& \tan \left(\frac{A-B}{2}\right)\end{array}$
Hence, option B is correct.