1. **Stating the problem:** We want to find the expression for $\tan^{-1}(\sqrt{x-3})$ or understand its behavior. 2. **Understanding the function:** The function is the inverse tangent (arctan) of the square root of $x-3$. The domain of $\sqrt{x-3}$ requires $x-3 \geq 0$, so $x \geq 3$. 3. **Formula and properties:** The inverse tangent function $\tan^{-1}(y)$ returns an angle $\theta$ such that $\tan(\theta) = y$ and $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. 4. **Intermediate work:** Let $y = \sqrt{x-3}$. Then $\tan^{-1}(y)$ is the angle whose tangent is $y$. 5. **Explanation:** For any $x \geq 3$, $\sqrt{x-3}$ is non-negative, so $\tan^{-1}(\sqrt{x-3})$ returns an angle between $0$ and $\frac{\pi}{2}$ radians. 6. **Summary:** The expression $\tan^{-1}(\sqrt{x-3})$ is defined for $x \geq 3$ and gives the angle whose tangent is $\sqrt{x-3}$. No further simplification is possible without additional context or equations.