How do you simplify the expression $\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right)$?

Hint:We know that arctan is nothing but the inverse of the tangent function. Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios which we will do in this question.

Complete step by step answer:
We need to simplify the term $\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right)$.
As this is an inverse trigonometric function, its value must provide an angle. Let this angle be $\theta $.
Therefore, we can write that
$\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right) = \theta $
We know that the range of inverse of tangent function is from $ - \dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$
which is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.
Therefore we can say that the angle $\theta $ belongs to this range $\left( { - \dfrac{\pi }{2},\dfrac{\pi
}{2}} \right)$.
We have, $\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right) = \theta $. Therefore, we can write $\tan \theta
= \dfrac{1}{{\sqrt 3 }}$.
Here, the value of $\tan \theta $ is greater than zero. This means that the angle $\theta $ does not belong to the range $\left( { - \dfrac{\pi }{2},0} \right)$.
$ \Rightarrow \theta \in \left( {0,\dfrac{\pi }{2}} \right)$
Thus, we can say that the angle $\theta $ is in the first quadrant..
We know that $\tan \theta = \dfrac{1}{{\sqrt 3 }}$ when the value of angle $\theta $ is $\dfrac{\pi
}{6}$.
Therefore, it is clear that $\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right) = \dfrac{\pi }{6}$.
Hence, by simplifying the expression $\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right)$, we get $\dfrac{\pi }{6}$ as our final answer.

Note:
In this type of question, the most important part is to consider the range of the inverse tangent function. Because, without considering it, we cannot obtain the single value of the function. Because of the range of the function and given value of ratio, we can know the quadrant of the angle. For example, here we have considered the range of inverse of tangent function which is$\left( { - \dfrac{\pi
}{2},\dfrac{\pi }{2}} \right)$.