Answer
$-\dfrac{\pi}{3}$
Work Step by Step
Suppose $\theta =\tan (-\dfrac{\pi}{3})$
This gives: $\tan^{-1} \theta=\tan^{-1}[\tan(-\dfrac{\pi}{3})]$
The inverse property states that $\tan^{-1}(\tan a)=a $ satisfies for every $ a $ in the range $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ . But the value $ a=\dfrac{2 \pi}{3}$ does not lie in the range of $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$.
Therefore, we have $\tan^{-1}[\tan(\dfrac{2\pi}{3})]=-\dfrac{\pi}{3}$
