Q. If $\tan^{- 1}{a + \tan^{- 1}{b = \sin^{- 1}{1 - \tan^{- 1}c}}}$, then
A) $a + b + c = abc$
B) $ab + bc + ca = abc$
C) $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{abc} = 0$
D) $ab + bc + ca = a + b + c$
Explanation
(c)
Given, $\tan^{- 1}{a + \tan^{- 1}{b + \tan^{- 1}{c = \sin^{- 1}1}}}$
$\therefore\ \tan^{- 1}{\left( \frac{a + b + c - abc}{1 - ab - bc - ca} \right) = \frac{\pi}{2}}$
$\Rightarrow \ \ \ \frac{a + b + c - abc}{1 - ab - bc - ca} = \frac{1}{0} \Rightarrow ab + bc + ca - 1 = 0$
$\Rightarrow \ \ \ \ \ \ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{abc} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.