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