1. **State the problem:** Find the value of $\tan \theta$ given the point $(-\sqrt{33}, -4)$ on the terminal side of angle $\theta$. 2. **Recall the formula:** The tangent of an angle $\theta$ in the coordinate plane is given by $$\tan \theta = \frac{y}{x}$$ where $(x,y)$ are the coordinates of the point on the terminal side of the angle. 3. **Identify coordinates:** Here, $x = -\sqrt{33}$ and $y = -4$. 4. **Calculate $\tan \theta$:** $$\tan \theta = \frac{-4}{-\sqrt{33}} = \frac{-4}{-\sqrt{33}}$$ 5. **Simplify the fraction:** Both numerator and denominator are negative, so they cancel out: $$\tan \theta = \frac{\cancel{-4}}{\cancel{-\sqrt{33}}} = \frac{4}{\sqrt{33}}$$ 6. **Rationalize the denominator:** Multiply numerator and denominator by $\sqrt{33}$: $$\tan \theta = \frac{4}{\sqrt{33}} \times \frac{\sqrt{33}}{\sqrt{33}} = \frac{4\sqrt{33}}{33}$$ 7. **Final answer:** $$\tan \theta = \frac{4\sqrt{33}}{33}$$ Since the point is in the third quadrant where both $x$ and $y$ are negative, $\tan \theta$ is positive, so the answer is $\boxed{\frac{4\sqrt{33}}{33}}$. Among the given options, the correct one is $\tan \theta = \frac{4\sqrt{33}}{33}$, but with a positive sign. **Note:** The option $\tan \theta = -\frac{4\sqrt{33}}{33}$ is incorrect because tangent is positive in the third quadrant. Hence, the correct value is $\tan \theta = \frac{4\sqrt{33}}{33}$.