1. **Problem Statement:** Find the slope of the tangent line to the graph of $f(x) = \tan(x)$ at $x = \frac{\pi}{6}$. 2. **Formula:** The slope of the tangent line to a function at a point is given by the derivative evaluated at that point. For $f(x) = \tan(x)$, the derivative is $$f'(x) = \sec^2(x)$$ 3. **Evaluate the derivative at $x = \frac{\pi}{6}$:** $$f'\left(\frac{\pi}{6}\right) = \sec^2\left(\frac{\pi}{6}\right)$$ Recall that $\sec(x) = \frac{1}{\cos(x)}$. 4. **Calculate $\cos\left(\frac{\pi}{6}\right)$:** $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$ 5. **Calculate $\sec\left(\frac{\pi}{6}\right)$:** $$\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$$ 6. **Calculate $f'\left(\frac{\pi}{6}\right)$:** $$f'\left(\frac{\pi}{6}\right) = \sec^2\left(\frac{\pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$$ 7. **Answer:** The slope of the tangent line at $x = \frac{\pi}{6}$ is $\frac{4}{3}$. This corresponds to option D. **Summary:** The slope of the tangent line to $f(x) = \tan(x)$ at $x = \frac{\pi}{6}$ is $\boxed{\frac{4}{3}}$.