Prove the given trigonometric expression ${{\cos }^{-1}}(x)+{{\cos }^{-1}}\left[ \dfrac{x}{2}+\dfrac{\sqrt{3-3{{x}^{2}}}}{2} \right]=\dfrac{\pi }{3}$

Hint: Substitute $x=\cos \theta $ and then use the identity: $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $, to simplify the second term. Write, $\dfrac{x}{2}+\dfrac{\sqrt{3-3{{x}^{2}}}}{2}=\dfrac{1}{2}x+\dfrac{\sqrt{3}}{2}\sqrt{1-{{x}^{2}}}$ and substitute, $\dfrac{1}{2}=\cos \dfrac{\pi }{3}$. Apply the formula: $\cos a\cos b+\sin a\sin b=\cos \left( a-b \right)$. Finally, use the identity: ${{\cos }^{-1}}(\cos x)=x$ to get the answer.

Complete step-by-step solution -
We have to prove: ${{\cos }^{-1}}(x)+{{\cos }^{-1}}\left[ \dfrac{x}{2}+\dfrac{\sqrt{3-3{{x}^{2}}}}{2} \right]=\dfrac{\pi }{3}$
$L.H.S={{\cos }^{-1}}(x)+{{\cos }^{-1}}\left[ \dfrac{x}{2}+\dfrac{\sqrt{3-3{{x}^{2}}}}{2} \right]$
Substituting, $x=\cos \theta $, we get,
$\begin{align}
  & L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{\cos \theta }{2}+\dfrac{\sqrt{3-3{{\cos }^{2}}\theta }}{2} \right] \\
 & \Rightarrow L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{\cos \theta }{2}+\dfrac{\sqrt{3(1-{{\cos }^{2}}\theta )}}{2} \right] \\
\end{align}$
Using the formula: $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $, we get,
$\begin{align}
  & L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{\cos \theta }{2}+\dfrac{\sqrt{3{{\sin }^{2}}\theta }}{2} \right] \\
 & \Rightarrow L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{\cos \theta }{2}+\dfrac{\sqrt{3}\left| \sin \theta \right|}{2} \right] \\
\end{align}$
Now, since we have assumed, $x=\cos \theta $, therefore, $\theta ={{\cos }^{-1}}x$ and the range of $\theta $ is from 0 to $\pi $. That means $\theta $ lies between the 1st and 2nd quadrant.
We know that sine of an angle is positive if the angle lies between 1st and 2nd quadrant. Therefore, the expression inside the mod is positive. So, removing the modulus we get,
$L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{\cos \theta }{2}+\dfrac{\sqrt{3}\sin \theta }{2} \right]$
The above expression can be written as:
\[\begin{align}
  & L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{\cos \theta }{2}+\dfrac{\sqrt{3{{\sin }^{2}}\theta }}{2} \right] \\
 & \Rightarrow L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \dfrac{1}{2}\cos \theta +\dfrac{\sqrt{3}}{2}\sin \theta \right] \\
\end{align}\]
We know that, $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$. Therefore, substituting $\cos \dfrac{\pi }{3}$ and $\sin \dfrac{\pi }{3}$ in place of $\dfrac{1}{2}$ and $\dfrac{\sqrt{3}}{2}$ respectively, we get,
\[L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \cos \dfrac{\pi }{3}\cos \theta +\sin \dfrac{\pi }{3}\sin \theta \right]\]
Applying the formula: $\cos a\cos b+\sin a\sin b=\cos \left( a-b \right)$, we get,
\[L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \cos \left( \dfrac{\pi }{3}-\theta \right) \right]\]
Now, using the identity: ${{\cos }^{-1}}(\cos x)=x$, we get,
\[\begin{align}
  & L.H.S=\theta +\left( \dfrac{\pi }{3}-\theta \right) \\
 & =\theta +\dfrac{\pi }{3}-\theta \\
 & =\dfrac{\pi }{3} \\
 & =R.H.S \\
\end{align}\]

Note: One may note that while removing the mod, we have to be careful about the certain conditions, that is, the range of ‘x’. Positive or negative value of a trigonometric function depends on the quadrant in which the angle is lying. In the above question, the angle $\theta $ lies in the first or second quadrant, therefore, the value of the trigonometric function inside the mod is positive.