Answer
$-\dfrac{3\pi}{8}$
Work Step by Step
We know that $f^{-1}(f(x))=x$.
Hence,
$\tan^{-1}\left(\tan(\frac{-3\pi}{8})\right)=\frac{-3\pi}{8}$
We also need that $ \frac{-\pi}{2} \leq x\leq \frac{\pi}{2}$, which is true here.
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