Step 1: Concept
$\sin^{-1}(\sin \theta) = \theta$ only if $\theta \in [-\pi/2, \pi/2]$. Otherwise, we must find an angle in the principal range.
Step 2: Analysis
$10$ radians is approximately $10 \times 57.3^{\circ} \approx 573^{\circ}$.
We know $3\pi \approx 9.42$. Thus, $10$ is near $3\pi$.
$\sin(10) = \sin(10 - 3\pi)$ is not correct because $\sin(3\pi - x) = \sin x$.
$\sin(10) = \sin(3\pi - 10)$.
Step 3: Conclusion
$3\pi - 10 \approx 9.42 - 10 = -0.58$, which lies within $[-\pi/2, \pi/2]$.
So, $\sin^{-1}(\sin 10) = 3\pi - 10$.
Final Answer: (C)