The mathematical expression in the image is:

$\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$

Solution:

This expression involves the inverse tangent (arctangent) function, which gives the angle whose tangent is the specified value.

1. Recall the trigonometric identity: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

2. Find the angle $\theta$ such that: $\tan(\theta) = \frac{\sqrt{3}}{3}$

3. From the unit circle or standard trigonometric values: $\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$

Thus: $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$

Final Answer:

$\boxed{\frac{\pi}{6}}$

Let me know if you need further explanation or additional steps!

Follow-up Questions:

1. What is the relationship between $\tan^{-1}$ and $\tan$?
2. Can you explain why $\tan^{-1}(x)$ is restricted to specific intervals?
3. How would you solve $\tan^{-1}(-\frac{\sqrt{3}}{3})$?
4. What is the value of $\sin\left(\frac{\pi}{6}\right)$ and $\cos\left(\frac{\pi}{6}\right)$?
5. How do inverse trigonometric functions apply to real-world problems?

Tip: Always check standard trigonometric values for angles like $\frac{\pi}{6}$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$—they simplify many calculations!