Inverse tangent in relation to sine and cosine

If ${\tan\theta}=\frac{\sin\theta}{\cos\theta}$, what would the function be if instead of $\tan\theta$ it were $\tan^{-1}\theta$?

I'm asking this because I am unsure of an explanation I saw in this video (at the given timecode), where he is showing the relationship between the $m$ the slope of a line, and $\tan\theta$. His example shows how $m=\tan\theta$, and how if $m = -1$, then $\tan^{-1}(-1)=\theta$. That process I understand. Afterwards, he asks if instead $m$ = $\frac{1}{2}$. Unless I'm going about it wrong, I think that the process is different with one-half.

I believe this question is asking something similar, although rather than that given answer, I am thinking of the answer can be shown with the unit circle.

If $x$ is a number then $\tan^{-1}(x)$ means the angle between $-\frac{\pi}{2}$ and $+\frac{\pi}{2}$ whose tangent is $x$.
For an angle in standard position (vertex at the origin, initial side the $x$-axis) the tangent of the angle equals the slope of the terminal side of the angle. This is true for all angles, not just those on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
So $\tan^{-1}\left(\frac{1}{2}\right)$ means the angle in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ whose terminal side has slope $\left(\frac{1}{2}\right)$.