If $y = \tan^{-1}\sqrt{\dfrac{1 + \sin 2x}{1 - \sin 2x}}$, then $\dfrac{\text{d}y}{\text{d}x}$ at $x = \dfrac{\pi}{6}$ is

We have $y = \tan^{-1}\sqrt{\dfrac{1 + \sin 2x}{1 - \sin 2x}}$.

Using the trigonometric identities $1 + \sin 2x = (\cos x + \sin x)^2$ and $1 - \sin 2x = (\cos x - \sin x)^2$, we get:

$y = \tan^{-1}\sqrt{\dfrac{(\cos x + \sin x)^2}{(\cos x - \sin x)^2}} = \tan^{-1}\left| \dfrac{\cos x + \sin x}{\cos x - \sin x} \right|$

At $x = \dfrac{\pi}{6}$, $\cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$ and $\sin \dfrac{\pi}{6} = \dfrac{1}{2}$. Since $\cos x > \sin x$, the expression inside the modulus is positive.

$y = \tan^{-1}\left( \dfrac{\cos x + \sin x}{\cos x - \sin x} \right)$

Dividing the numerator and the denominator by $\cos x$:

$y = \tan^{-1}\left( \dfrac{1 + \tan x}{1 - \tan x} \right)$

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

For $x = \dfrac{\pi}{6}$, $\dfrac{\pi}{4} + x = \dfrac{5\pi}{12}$, which lies in the principal branch $\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$.

$y = \dfrac{\pi}{4} + x$

Differentiating with respect to $x$:

$\dfrac{\text{d}y}{\text{d}x} = 1$

Answer: $1$