Q. If $\tan{A = 2\tan{B + \cot B}}$, then $2\tan{(A - B)}$ is equal to
A) $\tan B$
B) $2\tan B$
C) $\cot B$
D) $2\cot B$
Explanation
(c)
$2\tan{(A - B) = 2\left( \frac{\tan{A - \tan B}}{1 + \tan{A\tan B}} \right)}$
$= 2\left( \frac{2\tan{B + \cot{B - \tan B}}}{1 + \left( 2\tan{B + \cot B} \right)\tan B} \right)$
$\lbrack\because\tan{A = 2\tan B + \cot B}\rbrack$
$= \frac{2(\tan{B + \cot B})}{2(1 + \tan^{2}B)} = \cot B$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.