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)
Given that,
$\tan{A = 2\tan{B + \cot B}}$ ….(i)
Now$,\ 2\tan{(A - B) = 2\left( \frac{\tan{A - \tan B}}{1 + \tan{A\tan B}} \right)}$
$= 2\frac{(2\tan{B + \cot{B - \tan{B)}}}}{1 + (2\tan{B + \cot{B)\tan B}}}\ \$[from Eq.(i)]
$= 2\frac{\tan{B + \cot B}}{2(1 + \tan^{2}{B)}} = \frac{\cot{B(\tan^{2}{B + 1)}}}{(1 + \tan^{2}{B)}} = \cot B$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.