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To find the equation of the tangent line to the graph of \( y = \tan \theta \sec \theta \) at \( \theta = \frac{\pi}{4} \), we need to follow these steps:

### Step 1: Differentiate the given function
The equation is \( y = \tan \theta \sec \theta \). To find the slope of the tangent line, we need to compute the derivative of \( y \) with respect to \( \theta \).

We apply the product rule to differentiate \( y = \tan \theta \sec \theta \).

\[
\frac{dy}{d\theta} = \frac{d}{d\theta} (\tan \theta \sec \theta)
\]
Using the product rule, we get:
\[
\frac{dy}{d\theta} = \sec \theta \cdot \frac{d}{d\theta} (\tan \theta) + \tan \theta \cdot \frac{d}{d\theta} (\sec \theta)
\]

We know that:
\[
\frac{d}{d\theta} (\tan \theta) = \sec^2 \theta \quad \text{and} \quad \frac{d}{d\theta} (\sec \theta) = \sec \theta \tan \theta
\]

Substituting these, we get:
\[
\frac{dy}{d\theta} = \sec \theta \cdot \sec^2 \theta + \tan \theta \cdot (\sec \theta \tan \theta)
\]
\[
\frac{dy}{d\theta} = \sec^3 \theta + \sec \theta \tan^2 \theta
\]

### Step 2: Evaluate the derivative at \( \theta = \frac{\pi}{4} \)
At \( \theta = \frac{\pi}{4} \), we know:
\[
\sec \left( \frac{\pi}{4} \right) = \sqrt{2} \quad \text{and} \quad \tan \left( \frac{\pi}{4} \right) = 1
\]

Substitute these values into the derivative:
\[
\frac{dy}{d\theta} = (\sqrt{2})^3 + \sqrt{2} \cdot (1)^2
\]
\[
\frac{dy}{d\theta} = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2}
\]

Thus, the slope of the tangent line is \( 3\sqrt{2} \).

### Step 3: Find the point on the graph at \( \theta = \frac{\pi}{4} \)
We need the value of \( y \) when \( \theta = \frac{\pi}{4} \):
\[
y = \tan \left( \frac{\pi}{4} \right) \sec \left( \frac{\pi}{4} \right)
\]
\[
y = 1 \cdot \sqrt{2} = \sqrt{2}
\]

So, the point of tangency is \( \left( \frac{\pi}{4}, \sqrt{2} \right) \).

### Step 4: Equation of the tangent line
The equation of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency.

Here, \( m = 3\sqrt{2} \), \( x_1 = \frac{\pi}{4} \), and \( y_1 = \sqrt{2} \).

Substituting these values:
\[
y - \sqrt{2} = 3\sqrt{2} \left( \theta - \frac{\pi}{4} \right)
\]

This is the equation of the tangent line.

Would you like more details on any specific part, or have any questions?

Here are 5 related questions:
1. How would the derivative change if the function was \( y = \sec \theta \tan^2 \theta \)?
2. Can you explain why the product rule is necessary in this problem?
3. What if the point of tangency was at a different angle, such as \( \theta = \frac{\pi}{3} \)?
4. How would the tangent line equation change if the slope were negative?
5. How does the behavior of \( \sec \theta \) and \( \tan \theta \) influence the graph of the original function?

**Tip**: When differentiating products of trigonometric functions, always apply the product rule and simplify using known identities where possible.

Find the tangent line to the graph of y = tan(θ) sec(θ) at θ = π/4.

Learn how to find the tangent line to the graph of y = tan(θ) sec(θ) at θ = π/4. This calculus problem involves using the product rule for differentiation, trigonometric identities, and the equation of a tangent line. Step-by-step solutions for advanced high school students.

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