Answer
$\dfrac{ \pi}{3}$
Work Step by Step
Value of certain composite functions formula:
$f^{-1}(f(x)= \tan^{-1}{(\tan{x})} = x \hspace{25pt} -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$
Note that $-\dfrac{2 \pi}{3} \text{ isn't in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$.
Recall that"
$\tan{\theta} = \tan{(\theta + \pi)}$
Thus,
$\tan{\left(-\dfrac{2 \pi}{3} \right)} = \tan{\left(-\dfrac{2 \pi}{3} + \pi \right) }= \tan{\left(\dfrac{\pi}{3} \right)}$
Since $\dfrac{\pi}{3} \text{ is in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$, then
$\tan^{-1} {\left[\tan{\left(\dfrac{ \pi}{3} \right)} \right]} = \dfrac{ \pi}{3}$
Therefore,
$\tan^{-1} {\left[\tan{\left(-\dfrac{ 2\pi}{3} \right)} \right]} = \boxed{\dfrac{ \pi}{3}}$
