Q. If $A$=$\tan^{- 1}\left( \frac{x\sqrt{3}}{2k - x} \right)\$and $B = \tan^{- 1}{\left( \frac{2x - k}{k\sqrt{3}} \right),}$then the value of $A - B$ is

A) $10{^\circ}$
B) 45${^\circ}$
C) $60{^\circ}$
D) $30{^\circ}$
Explanation

(d)
We know that,
$\tan{(A - B) = \frac{\tan{A - \tan B}}{1 + \tan{A\tan B}}}$
$= \frac{\frac{x\sqrt{3}}{2k - x} - \frac{2x - k}{k\sqrt{3}}}{1 + \frac{x\sqrt{3}}{2k - x}.\frac{2x - k}{k\sqrt{3}}} = \frac{1}{\sqrt{3}}$
$\Rightarrow A - B = 30{^\circ}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.