Q. $2\tan^{- 1}{\left( \frac{1}{3} \right) + \tan^{- 1}\left( \frac{1}{7} \right)}$is equal to

A) $\left( \frac{49}{29} \right)$
B) $\frac{\pi}{2}$
C) $- \left( \frac{49}{29} \right)$
D) $\frac{\pi}{4}$
Explanation

(d)
$2\tan^{- 1}{\left( \frac{1}{3} \right) + \tan^{- 1}\left( \frac{1}{7} \right)}$
$= \tan^{- 1}{\left\lbrack \frac{2\left( \frac{1}{3} \right)}{1 - \frac{1}{9}} \right\rbrack + \tan^{- 1}\left( \frac{1}{7} \right)}$
$= \tan^{- 1}{\left( \frac{3}{4} \right) + \tan^{- 1}\left( \frac{1}{7} \right)}$
$= \tan^{- 1}\left( \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \times \frac{1}{7}} \right)$
=$\tan^{- 1}{\left( \frac{25}{25} \right) = \frac{\pi}{4}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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