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Solution

For $\tan^{-1}(\tan \theta)$, the result lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $13$ radians is beyond this interval, $\tan 13 = \tan(13 - 4\pi)$ $\Rightarrow \tan^{-1}(\tan 13) = 13 - 4\pi$