Q. The relation $\tan^{- 1}{\left( \frac{1 + x}{1 - x} \right) = \frac{\pi}{4} + \tan^{- 1}x}$ holds true for all
A) $x \in R$
B) $x \in ( - \infty,\ 1)$
C) $x \in ( - 1,\ \infty)$
D) $x \in ( - \infty,\ - 1)$
Explanation
(b)
Given, $\tan^{- 1}{\left( \frac{1 + x}{1 - x} \right) = \frac{\pi}{4} + \tan^{- 1}x}$
RHS$= \frac{\pi}{4} + \tan^{- 1}{x = \tan^{- 1}{1 + \tan^{- 1}x}}$
$\ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{1 + x}{1 - x} \right)$, if $x < 1$
$\therefore\ \ \ \ \ \ x \in ( - \infty,\ 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.