How do you evaluate $\tan \left( {\dfrac{{5\pi }}{6}} \right)$ ?

Hint: The value of tangent function can be find by rewriting the given function that is $\tan \left( {\dfrac{{5\pi }}{6}} \right)$ as $\tan \left( {\pi - \dfrac{\pi }{6}} \right)$ . The function $\tan \left( {\pi - \dfrac{\pi }{6}} \right)$ is in the form of $\tan (A - B)$ which is given by: $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 - \tan A\tan B}}$ so now by substituting $A = \pi $ and $B = \dfrac{\pi }{6}$ and simplifying the expression we get required answer.

Complete step by step answer:
The given function is $\tan \left( {\dfrac{{5\pi }}{6}} \right)$ for which if we do not have the direct values in the standard trigonometric ratio table, we can do it as below.
Now we need to rewrite the function $\tan \left( {\dfrac{{5\pi }}{6}} \right)$ as $\tan \left( {\pi - \dfrac{\pi }{6}} \right)$ , which makes simplification easier. The function $\tan \left( {\pi - \dfrac{\pi }{6}} \right)$ is in the form of $\tan (A - B)$. The $\tan (A - B)$ formula is given by: $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 - \tan A\tan B}}$ . Now we can use this formula to find the required answer.
Therefore, now by substituting $A = \pi $ and $B = \dfrac{\pi }{6}$ in the above given formula, we get
$\tan \left( {\dfrac{{5\pi }}{6}} \right) = \tan (\pi - \dfrac{\pi }{6}) = \dfrac{{\tan \pi - \tan \left( {\dfrac{\pi }{6}} \right)}}{{1 - \tan \pi \tan \left( {\dfrac{\pi }{6}} \right)}}$
Now, substitute the standard values of the tangent function in the above expression for simplification purposes.
We know that the values for the above functions which is given by:
 $\
  \tan \pi = 0 \\
  \tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }} \\
\ $
Now substitute these values in the above expression for simplification purpose, we get
$ \Rightarrow \tan \left( {\dfrac{{5\pi }}{6}} \right) = \tan (\pi - \dfrac{\pi }{6}) = \dfrac{{0 - \dfrac{1}{{\sqrt 3 }}}}{{1 - 0 \times \dfrac{1}{{\sqrt 3 }}}}$
On simplifying the above expression, we get
$ \Rightarrow \tan \left( {\dfrac{{5\pi }}{6}} \right) = \tan (\pi - \dfrac{\pi }{6}) = \dfrac{{ - \dfrac{1}{{\sqrt 3 }}}}{{1 - 0}}$
$ \Rightarrow \tan \left( {\dfrac{{5\pi }}{6}} \right) = \tan (\pi - \dfrac{\pi }{6}) = - \dfrac{1}{{\sqrt 3 }}$

Therefore, the value of given function $\tan \left( {\dfrac{{5\pi }}{6}} \right)$ is $ - \dfrac{1}{{\sqrt 3 }}$.

Note:
The given problem can be solved in another way that is by using quadrants. The given function that is $\tan \left( {\dfrac{{5\pi }}{6}} \right)$ where the angle $\dfrac{{5\pi }}{6}$ that is 150 degrees lies in between 90 degree and the 180 degree which means the given function lies in second quadrant where tangent function is negative. Therefore we can write as $\tan {150^ \circ } = \tan ({180^ \circ } - {30^ \circ })$ here we know that $\tan (180 - \theta ) = - \tan \theta $ . Hence we can write as $\tan 150 = - \tan 30 = - \dfrac{1}{{\sqrt 3 }}$. If you do not remember the formula then you can use this process. Or you can use a calculator to find this.