If we have $2y={{\left( {{\cot }^{-1}}\left( \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right)}^{2}}$ then the value of $\dfrac{dy}{dx}$ is equal to.
$\begin{align}
  & a)x-\dfrac{\pi }{6} \\
 & b)x+\dfrac{\pi }{6} \\
 & c)2x-\dfrac{\pi }{6} \\
 & d)2x-\dfrac{\pi }{3} \\
\end{align}$

Hint: Now fist we will consider $\dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}$ and divide the equation by cosx. Now we know that $\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$ hence using this we will get the expression in form of tan. Now we will use the formula $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . Then we will convert $\tan \theta $ into $\cot \theta $ and differentiate the expression with the formula $\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right)$ .

Complete answer:
Now we are given with the expression $2y={{\left( {{\cot }^{-1}}\left( \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right)}^{2}}$ .
Now first let us consider the expression $\dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}$ .
We will first try to simplify this expression.
Now let us divide the expression by cosx. Hence we get,
$\dfrac{\sqrt{3}+\dfrac{\sin x}{\cos x}}{1-\sqrt{3}\dfrac{\sin x}{\cos x}}=\dfrac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}$
Now we know that the value of $\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$ . Hence substituting this in the equation above we get
\[\dfrac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}=\dfrac{\tan \left( \dfrac{\pi }{3} \right)+\tan x}{1-\tan \left( \dfrac{\pi }{3} \right)\tan x}\]
Now we know that $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ and the equation above is in this form. Hence using this formula we can say
\[\dfrac{\tan \left( \dfrac{\pi }{3} \right)+\tan x}{1-\tan \left( \dfrac{\pi }{3} \right)\tan x}=\tan \left( \dfrac{\pi }{3}+x \right)\]
Hence we now have $\dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}=\tan \left( \dfrac{\pi }{3}+x \right)......................\left( 1 \right)$ .
Now we want to convert this into cot. We know that the relation between tan and cot is $\cot \left( \dfrac{\pi }{2}-x \right)=\tan x$
Hence we have $\cot \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{3}+x \right) \right)=\tan \left( \dfrac{\pi }{3}+x \right)$ using this substitution in equation (1) we get.
$\dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}=\cot \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{3}+x \right) \right)$
Now if we take ${{\cot }^{-1}}$ on both sides we get.
$\begin{align}
  & {{\cot }^{-1}}\left( \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right)={{\cot }^{-1}}\left( \cot \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{3}+x \right) \right) \right) \\
 & \Rightarrow {{\cot }^{-1}}\left( \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right)=\dfrac{\pi }{2}-\left( \dfrac{\pi }{3}+x \right) \\
\end{align}$
Now let us square on both sides.
\[{{\left( {{\cot }^{-1}}\left( \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right)}^{2}}={{\left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{3}+x \right) \right)}^{2}}\]
But we have $2y={{\left( {{\cot }^{-1}}\left( \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right)}^{2}}$
This means $2y={{\left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{3}+x \right) \right)}^{2}}$
Hence we get $2y={{\left( \dfrac{\pi }{2}-\dfrac{\pi }{3}-x \right)}^{2}}......................(2)$
Now we know that according to chain rule $\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right)$
Using this formula let us differentiate equation (2).
$\begin{align}
  & 2\dfrac{dy}{dx}=2\left( \dfrac{\pi }{2}-\dfrac{\pi }{3}-x \right)\left( -1 \right) \\
 & \Rightarrow \dfrac{dy}{dx}=-\left( \dfrac{\pi }{2}-\dfrac{\pi }{3}-x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=-\left( \dfrac{3\pi -2\pi }{6}-x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=-\left( \dfrac{\pi }{6}-x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=x-\dfrac{\pi }{6} \\
\end{align}$
Hence option a is the correct option.

Note:
Now since we know the chain rule for differentiation we shouldn’t directly start with this in these types of questionss. It’s always better to simplify first. Especially if there is an inverse trigonometric function with trigonometric ratios.