Question

If $sinA=nsinB$ , then $\frac{n-1}{n+1}tan\frac{A+B}{2}$ is equal to

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

Answer 1

Answer: (B)

Given, $\sin A=n\sin B$
$\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B}$
Applying componendo and dividendo, we get
$\frac{n-1}{n+1}=\frac{\sin A-\sin B}{\sin A+\sin B}$
$=\frac{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}$
$\Rightarrow \frac{n-1}{n+1}=\tan \left(\frac{A-B}{2}\right)\cot \left(\frac{A+B}{2}\right)$
$\Rightarrow \frac{n-1}{n+1}\tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{A-B}{2}\right)$