Answer
$-\dfrac{ \pi}{5}$
Work Step by Step
Value of certain composite functions formula:
$f^{-1}(f(x)= \sin^{-1}{(\sin{x})} = x \hspace{25pt} -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$
Note that $\dfrac{4 \pi}{5} \text{ isn't in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$.
Recall that:
$\tan{\theta} = \tan{(\theta - \pi)}$
Thus,
$\tan{\left(\dfrac{4 \pi}{5} \right)} = \tan{\left(\dfrac{4 \pi}{5} - \pi \right) }= \tan{\left(-\dfrac{\pi}{5} \right)}$
Since $-\dfrac{\pi}{5} \text{ is in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$, then
$\tan^{-1} {\left[\tan{\left(-\dfrac{ \pi}{5} \right)} \right]} = -\dfrac{ \pi}{5}$
Therefore,
$\tan^{-1} {\left[\tan{\left(\dfrac{ 4\pi}{5} \right)} \right]} = \boxed{-\dfrac{ \pi}{5}}$
