Q. If $\sin{A = nsin\ B}$, 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
Explanation
(b)
Given, $\sin{A = nsin\ B\ \ \ \Rightarrow \ \frac{n}{1} = \frac{\sin A}{\sin B}}$
Applying componendo and dividend, we get
$\frac{n - 1}{n + 1} = \frac{\sin{A - \sin B}}{\sin{A + \sin B}}$
$= \frac{2\cos{\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)}}{2\sin{\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}}$
$\Rightarrow \ \ \frac{n - 1}{n + 1} = \tan{\left( \frac{A - B}{2} \right)\cot\left( \frac{A + B}{2} \right)}$
$\Rightarrow \ \ \frac{n - 1}{n + 1}\tan{\left( \frac{A + B}{2} \right) = \tan\left( \frac{A - B}{2} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.