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The problem asks for the equation of the tangent line to the curve \( y = \tan^2(x) \) at the point \( \left(\frac{\pi}{3}, 3 \right) \). 

We need to:
1. Differentiate \( y = \tan^2(x) \) to find the slope of the tangent line at \( x = \frac{\pi}{3} \).
2. Use the point-slope form of a line to write the equation.

### Step 1: Find the derivative of \( y = \tan^2(x) \)

Using the chain rule:
\[
\frac{dy}{dx} = 2 \tan(x) \cdot \sec^2(x)
\]
This is the derivative of \( y = \tan^2(x) \).

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{3} \)

First, calculate \( \tan\left(\frac{\pi}{3}\right) \) and \( \sec\left(\frac{\pi}{3}\right) \):
\[
\tan\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \sec\left(\frac{\pi}{3}\right) = 2
\]
Now substitute into the derivative formula:
\[
\frac{dy}{dx} \Big|_{x = \frac{\pi}{3}} = 2 \cdot \sqrt{3} \cdot (2^2) = 2 \cdot \sqrt{3} \cdot 4 = 8\sqrt{3}
\]
So, the slope \( m = 8\sqrt{3} \).

### Step 3: Find the equation of the tangent line

We use the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = 8\sqrt{3} \), \( x_1 = \frac{\pi}{3} \), and \( y_1 = 3 \). Substituting these values:
\[
y - 3 = 8\sqrt{3} \left( x - \frac{\pi}{3} \right)
\]
Simplifying this into slope-intercept form \( y = mx + b \), we get:
\[
y = 8\sqrt{3}x - 8\sqrt{3}\frac{\pi}{3} + 3
\]
\[
y = 8\sqrt{3}x - \frac{8\pi\sqrt{3}}{3} + 3
\]
So, \( m = 8\sqrt{3} \) and \( b = 3 - \frac{8\pi\sqrt{3}}{3} \).

Let me know if you'd like further clarifications or additional steps.

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**Relative questions**:
1. How does the chain rule apply in differentiating \( \tan^2(x) \)?
2. What is the point-slope form of a line, and why do we use it for tangent lines?
3. Why is the derivative of \( \tan(x) \) related to \( \sec^2(x) \)?
4. How can we simplify terms involving \( \pi \) in the final answer for \( b \)?
5. What would happen if we changed the function to \( \cot(x) \) instead of \( \tan(x) \)?

**Tip**: When solving problems involving the tangent line, always start by differentiating to find the slope before applying the point-slope formula.

Find an equation of the tangent line to the curve y = tan^2(x) at the point (π/3, 3).

Learn how to find the equation of the tangent line to the curve y = tan^2(x) at the point (π/3, 3). This problem involves applying the chain rule, evaluating trigonometric functions, and using the point-slope form of a line. Ideal for students in Calculus.

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