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**Hint:**We start solving the problem by recalling the principal value or range of the inverse cosine function ${{\cos }^{-1}}\left( x \right)$. We then find the value of the $\cos \left( \dfrac{7\pi }{6} \right)$ using the fact that $\cos \left( \pi +\theta \right)=-\cos \theta $ and substitute it in the given ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$. We then follow the range of inverse cosine function ${{\cos }^{-1}}\left( x \right)$ to find the value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$.

**Complete step by step answer:**

According to the problem, we need to find the value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$.

We know that the function ${{\cos }^{-1}}\left( x \right)$ has principal value or the range of ${{\cos }^{-1}}\left( x \right)$ is restricted between 0 and $\pi $. This means that ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$ has value between 0 and $\pi $ -(1).

Let us solve for the value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$.

Now let us find the value of $\cos \left( \dfrac{7\pi }{6} \right)$. We know that $\cos \left( \pi +\theta \right)=-\cos \theta $.

$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)=\cos \left( \pi +\dfrac{\pi }{6} \right)$.

$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)=-\cos \left( \dfrac{\pi }{6} \right)$.

$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)=\dfrac{-\sqrt{3}}{2}$.

So, we have got ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)={{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)$.

We know that ${{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)=\dfrac{5\pi }{6}$, following the definition of the principal value of ${{\cos }^{-1}}\left( x \right)$ as mentioned in equation (1).

So, we get ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)=\dfrac{5\pi }{6}$.

We have found the value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$ as $\dfrac{5\pi }{6}$.

**So, the correct answer is “Option b”.**

**Note:**We should not directly say the value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$ as $\dfrac{7\pi }{6}$. This is because of the fact that the value $\dfrac{7\pi }{6}$ is not in the principal value of the function ${{\cos }^{-1}}\left( x \right)$. Whenever we get this type of problem, we should answer it in between principal values. If it is specified that the values outside the principal range are also allowed, then we can say all the possible answers. We don’t need to remember the hectic formulas related to inverse trigonometric functions while solving this type of problems.

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