How do you evaluate $\tan \left( { - \dfrac{{2\pi }}{3}} \right)$ ?

Hint: Here in this question they have asked to find the value of $\tan \left( { - \dfrac{{2\pi }}{3}} \right)$ . Now to simplify the given function, we can rewrite the tangent function as $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ , here $\theta = - \dfrac{{2\pi }}{3}$ . So by substituting the $\theta $ value in the above expression, we can arrive at the required answer.

Complete step by step answer:
In the given problem they have asked us to evaluate the function $\tan \left( { - \dfrac{{2\pi }}{3}} \right)$ . To make the simplification easier, we can write the tangent function as $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ . From the given function that is $\tan \left( { - \dfrac{{2\pi }}{3}} \right)$ , $\theta = - \dfrac{{2\pi }}{3}$ .
Now substitute the value of $\theta $ in the expression $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ , we get
$\tan \left( { - \dfrac{{2\pi }}{3}} \right) = \dfrac{{\sin \left( { - \dfrac{{2\pi }}{3}} \right)}}{{\cos \left( { - \dfrac{{2\pi }}{3}} \right)}}$
We know that in trigonometry we have $\sin \left( { - \theta } \right) = - \sin \theta $ and $\cos \left( { - \theta } \right) = \cos \theta $ . Therefore, the above expression becomes
$ \Rightarrow \tan \left( { - \dfrac{{2\pi }}{3}} \right) = \dfrac{{ - \sin \left( {\dfrac{{2\pi }}{3}} \right)}}{{\cos \left( {\dfrac{{2\pi }}{3}} \right)}}$
Now, to find the required values we need to substitute the sine and cosine values. We know that $\sin \left( {\dfrac{{2\pi }}{3}} \right) = \dfrac{{\sqrt 3 }}{2}$ and $\cos \left( {\dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}$ . Now substitute these values in the above expression, we get
$ \Rightarrow \tan \left( { - \dfrac{{2\pi }}{3}} \right) = \dfrac{{ - \dfrac{{\sqrt 3 }}{2}}}{{ - \dfrac{1}{2}}}$
In the above expression we have negative signs both in the numerator and the denominator which gets cancelled. And the number 2 also gets cancelled upon simplification. Which is written as below
$ \Rightarrow \tan \left( { - \dfrac{{2\pi }}{3}} \right) = \dfrac{{\sqrt 3 }}{{{2}}} \times \dfrac{{{2}}}{1}$
 Therefore, we get
$ \Rightarrow \tan \left( { - \dfrac{{2\pi }}{3}} \right) = \sqrt 3 $

Hence the value of $\tan \left( { - \dfrac{{2\pi }}{3}} \right)$ is $\sqrt 3 $.

Note: Whenever we have this type of problems we need to know how to rewrite the given tangent function with respect to sine and cosine functions. Once we rewrite the given function we can directly substitute the corresponding standard values and simplify the expression to get the correct answer. If you use the standard values of tangent function, ten directly you can get the correct answer.