Considering only the principal values of inverse functions, the set...

Given $\tan^{- 1}\left(2x\right) + \tan^{- 1}\left(3x\right) = \dfrac{\pi}{4}$

$⇒ ta{n}^{- 1}\left(\dfrac{2x + 3x}{1 - 2x \times 3x}\right) = \dfrac{\pi}{4}$

Applying tangent on both sides,

$⇒ \dfrac{5x}{1 - 6{x}^{2}} = 1\left[∵ tan\dfrac{\pi}{4} = 1\right]$

$⇒ 5x = 1 - 6{x}^{2}$

$⇒ 6{x}^{2} + 5x - 1 = 0$

$⇒ 6{x}^{2} + 6x - x - 1 = 0$

$⇒ \left(6x - 1\right)\left(x + 1\right) = 0$

$⇒ x = - 1, \dfrac{1}{6}$

But, since $x \ge 0$.

Hence, $x = \dfrac{1}{6}$

$∴ A$ has only one element and thus, it is a singleton.