Q. If $\tan^{- 1}{(x + 2) + \tan^{- 1}{(x - 2) - \tan^{- 1}{\left( \frac{1}{2} \right) = 0},\ }}$ then one of the values of $x$ is equal to
A) $- 1$
B) 5
C) $\frac{1}{2}$
D) 1
Explanation
(d)
$\tan^{- 1}{(x + 2) + \tan^{- 1}{(x - 2) = \tan^{- 1}\frac{1}{2}}}$
$\Rightarrow \tan^{- 1}{\frac{x + 2 + x - 2}{1 - (x + 2)(x - 2)} = \tan^{- 1}\frac{1}{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \frac{2x}{1 - x^{2} + 4} = \frac{1}{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4x = 5 - x^{2}$
$\Rightarrow \ \ \ \ \ \ x^{2} + 4x - 5 = 0$
$\Rightarrow \ \ (x - 1)(x + 5) = 0$
$\Rightarrow \ \ x = 1,\ - 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.