The problem is asking for the exact value of $\sin^{-1} \left( \sin \left( \frac{5\pi}{6} \right) \right)$.

Step-by-Step Solution:

1. Understanding the inverse sine (arcsine):

   • The function $\sin^{-1}(x)$ (also called arcsine) returns the angle $y$ such that $\sin(y) = x$.
   • The principal range of $\sin^{-1}(x)$ is $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$. This is crucial because arcsine only returns values in this range.

2. Simplifying $\sin \left( \frac{5\pi}{6} \right)$:

   • $\frac{5\pi}{6}$ is in the second quadrant where sine is positive.
   • The reference angle for $\frac{5\pi}{6}$ is $\frac{\pi}{6}$, so $\sin \left( \frac{5\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$.

3. Finding $\sin^{-1} \left( \frac{1}{2} \right)$:

   • We now need to find the angle $\theta$ such that $\sin(\theta) = \frac{1}{2}$ and $\theta$ lies in the range $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.
   • The angle in this range is $\frac{\pi}{6}$.

Conclusion:

The exact value of $\sin^{-1} \left( \sin \left( \frac{5\pi}{6} \right) \right)$ is $\frac{\pi}{6}$.

So, the correct answer is: $\boxed{A} \quad \frac{\pi}{6}$

Would you like more details on inverse trigonometric functions?

Here are 5 related questions for further practice:

1. What is the value of $\sin^{-1} \left( \sin \left( \frac{7\pi}{6} \right) \right)$?
2. Simplify $\sin^{-1} \left( \sin \left( -\frac{3\pi}{4} \right) \right)$.
3. What is the principal value of $\cos^{-1} \left( \cos \left( \frac{4\pi}{3} \right) \right)$?
4. Find $\tan^{-1} \left( \tan \left( \frac{5\pi}{4} \right) \right)$.
5. What is the range of the inverse sine function?

Tip: When solving problems involving inverse trigonometric functions, always pay attention to the principal range of the inverse function.