The expression you provided involves inverse tangent (also known as arctangent). Let's break it down:

1. First Expression:

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

   We are looking for an angle whose tangent value is $\frac{\sqrt{3}}{3}$.

   Recall that:

   $\tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{or} \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$

   So, $\frac{\sqrt{3}}{3}$ is the same as $\frac{1}{\sqrt{3}}$. Therefore:

   $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \quad \text{or} \quad 30^\circ$

2. Second Expression:

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

   We are now looking for an angle whose tangent value is $-\frac{1}{\sqrt{3}}$.

   The tangent function is negative in the second and fourth quadrants. Since $\tan\left(30^\circ\right) = \frac{1}{\sqrt{3}}$, we know:

   $\tan(150^\circ) = -\frac{1}{\sqrt{3}} \quad \text{or} \quad \tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}$

   Therefore:

   $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \quad \text{or} \quad -30^\circ$

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So, while the expressions $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$ and $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ do not directly equal each other, they are related through symmetry of the tangent function in the unit circle.

Explanation of the relationship:

• $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$ yields an angle of $30^\circ$ or $\frac{\pi}{6}$.
• $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ yields an angle of $-30^\circ$ or $-\frac{\pi}{6}$.

Since the tangent function has symmetry about the origin, $\tan^{-1}(-x) = -\tan^{-1}(x)$. This explains why:

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

Questions for further exploration:

1. How does the inverse tangent behave for angles in different quadrants?
2. What is the relationship between the inverse trigonometric functions and the unit circle?
3. How can the symmetry of trigonometric functions help simplify expressions?
4. What is the principal value range for $\tan^{-1}(x)$?
5. How does the behavior of $\tan^{-1}(x)$ change when $x$ is positive versus negative?

Tip: When working with inverse trigonometric functions, remember that $\tan^{-1}(x)$ always returns the angle in the range $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, so angles must be chosen accordingly!