Q. If $\tan^{- 1}{x + \tan^{- 1}{y = \frac{\pi}{4}}}$, then
A) $x + y + xy = 1$
B) $x + y - xy = 1$
C) $x + y + xy + 1 = 0$
D) $x + y - xy + 1 = 0$
Explanation
(a)
Given, $\tan^{- 1}{x + \tan^{- 1}{y = \frac{\pi}{4}}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tan^{- 1}{\left( \frac{x + y}{1 - xy} \right) = \frac{\pi}{4}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{x + y}{1 - xy} = 1$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + y + xy = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.