If $$2y=\left(\cot^{-1}\left(\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)^{ }\right)^2$$, $$\forall x \in \left(0, \frac{\pi}{2}\right)$$, then $$\frac{dy}{dx}$$ is equal to:

$$2y = \left\{ \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \right\}^2$$

$$2y = \left\{ \cot^{-1} \left( \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} \right) \right\}^2$$

$$2y = \left\{ \cot^{-1} \left( \tan \left( \frac{\pi}{3} + x \right) \right) \right\}^2$$

$$2y = \left\{ \frac{\pi}{2} - \tan^{-1} \left( \tan \left( \frac{\pi}{3} + x \right) \right) \right\}^2 = \left( \frac{\pi}{2} - \left( \frac{\pi}{3} + x \right) \right)^2$$

$$2y = \left( x - \frac{\pi}{6} \right)^2$$

$$2y = x^2 - \frac{\pi}{3}x + \frac{\pi^2}{36}$$

$$y' = x - \frac{\pi}{6}$$

Hence The correct ans is Option C