Explanation

(a)
$\tan^{- 1}{2x + \tan^{- 1}{3x = \frac{\pi}{4}}}$
$\Rightarrow \frac{3x + 2x}{1 - 6x^{2}} = \frac{\pi}{4}$
$\Rightarrow 5x = 1 - 6x^{2}$
$\Rightarrow 6x^{2} + 5x - 1 = 0$
$\Rightarrow x = - 1,\frac{1}{6}$
But when $x = - 1,$
$\tan^{- 1}{2x = \tan^{- 1}{( - 2) < 0}}$
And $\tan^{- 1}{3x = \tan^{- 1}{( - 3) < 0}}$
This value will not satisfy the given equation
Hence, $x = \frac{1}{6}$