Always respect the principal value branches! For inverse sine, the output must fall between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. For inverse cotangent, the output must be strictly between $0$ and $\pi$.
Step 1: Split the problem.
We must find $\sin^{-1}[\sin(-600^\circ)]$ and $\cot^{-1}(-\sqrt{3})$, then add them.
Step 2: Simplify the angle inside sine.
Sine repeats every $360^\circ$. Add $720^\circ$ to $-600^\circ$.
\[ -600^\circ + 720^\circ = 120^\circ \]
So $\sin(-600^\circ) = \sin 120^\circ = \frac{\sqrt{3}}{2}$.
Step 3: Apply the inverse sine.
The answer of $\sin^{-1}$ must lie between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \]
Step 4: Handle the inverse cotangent.
The answer of $\cot^{-1}$ must lie between $0$ and $\pi$. Since the input is negative, the angle is in the second quadrant.
\[ \cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]
Step 5: Add the two results.
\[ \frac{\pi}{3} + \frac{5\pi}{6} = \frac{2\pi}{6} + \frac{5\pi}{6} = \frac{7\pi}{6} \]
Step 6: Conclusion.
Always respect the allowed ranges of inverse functions to get the right answer.
\[ \boxed{\frac{7\pi}{6} \text{ (Option 4)}} \]