Q. $\tan^{- 1}{\frac{x}{y} - \tan^{- 1}\frac{x - y}{x + y}}\$is equal to (where $x < y > 0)$
A) $- \frac{\pi}{4}$
B) $\frac{\pi}{4}$
C) $\frac{3\pi}{4}$
D) None of these
Explanation
(b)
$\tan^{- 1}{\frac{x}{y} - \tan^{- 1}\frac{x - y}{x + y}}$
$= \tan^{- 1}{\frac{x}{y} - \tan^{- 1}\left\lbrack \frac{1 - \frac{y}{x}}{1 + \frac{y}{x}} \right\rbrack}$
$= \tan^{- 1}{\frac{x}{y} - {\tan^{- 1}\ 1}{+ \tan^{- 1}\frac{y}{x}}}$
$= \tan^{- 1}{\frac{x}{y} + \cot^{- 1}{\frac{x}{y} - \tan^{- 1}1}}$
$= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.