Q. $\tan^{- 1}{\frac{m}{n} - \tan^{- 1}\frac{m - n}{m + n}}$ is equal to
A) $\tan^{- 1}\frac{n}{m}$
B) $\tan^{- 1}\frac{m + n}{m - n}$
C) $\frac{\pi}{4}$
D) $\tan^{- 1}\left( \frac{1}{2} \right)$
Explanation
(c)
$\tan^{- 1}{= \frac{m}{n} - \tan^{- 1}\frac{m - n}{m + n}}$
$\ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{\frac{m}{n} - \tan^{- 1}\frac{\frac{m}{n} - 1}{1 + \frac{m}{n}}}$
$\ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{\frac{m}{n} - \tan^{- 1}{\frac{m}{n} + \tan^{- 1}{(1) = \frac{\pi}{4}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.