Question

If $tan^{-1}(x - 1) + tan^{-1} x + tan^{-1}(x + 1)=tan^{-1}\, 3x$, then the values of $x$ are

• A. $\pm \frac{1}{2}$
• B. $0, \frac{1}{2}$
• C. $0, - \frac{1}{2}$
• D. $0, \pm \frac{1}{2}$

Answer 1

Answer: (D)

We have,
$tan^{-1}\left(x - 1\right) + tan^{-1} x + tan^{-1}\left(x +1\right)=tan^{-1}\, 3x$
$\Rightarrow tan^{-1}\left(x - 1\right) + tan^{-1} \left(x +1\right)$
$= tan^{-1}\, 3x - tan^{-1}\, x$
$\Rightarrow tan^{-1} \left\{\frac{\left(x-1\right)+\left(x+1\right)}{1-\left(x^{2}-1\right)}\right\}$
$= tan^{-1}\left\{\frac{3x-x}{1+3x^{2}}\right\}$
$\Rightarrow \frac{2x}{2 - x^{2}} = \frac{2x}{1 + 3x^{2}}$
$\Rightarrow 2x \left(1+3x^{2}\right) = 2x \left( 2 - x^{2}\right)$
$\Rightarrow 2x \left(4x^{2}-1\right) = 0$
$\Rightarrow x = 0$ or $x = \pm \frac{1}{2}$