To find the equation of the tangent line to the curve $y = \tan(x)$ at the point where $x = \frac{\pi}{3}$:

Step 1: Determine the point of tangency

We substitute $x = \frac{\pi}{3}$ into $y = \tan(x)$: $y = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}.$ Thus, the point of tangency is: $\left(\frac{\pi}{3}, \sqrt{3}\right).$

Step 2: Find the derivative

The derivative of $y = \tan(x)$ is: $\frac{dy}{dx} = \sec^2(x).$ At $x = \frac{\pi}{3}$: $\sec^2\left(\frac{\pi}{3}\right) = \left( \frac{2}{\sqrt{3}} \right)^2 = \frac{4}{3}.$

So the slope of the tangent line is: $m = \frac{4}{3}.$

Step 3: Write the equation of the tangent line

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1),$ where $(x_1, y_1) = \left(\frac{\pi}{3}, \sqrt{3}\right)$ and $m = \frac{4}{3}$, we get: $y - \sqrt{3} = \frac{4}{3}\left(x - \frac{\pi}{3}\right).$

Simplify: $y = \frac{4}{3}x - \frac{4}{9}\pi + \sqrt{3}.$

Final Answer:

The equation of the tangent line is: $y = \frac{4}{3}x - \frac{4}{9}\pi + \sqrt{3}.$

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Related Questions:

1. What is the slope of a tangent to $y = \tan(x)$ at $x = \pi/4$?
2. How do you compute $\sec^2(x)$ for other angles like $x = 0$ or $x = \pi/6$?
3. Can you derive the equation of a normal line instead of a tangent?