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.