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The problem from the image reads:

**Problem (10):**
Given:
\[
x = 5 + \sec 3\theta
\]
and
\[
y = 1 - \tan 3\theta,
\]
find \(\frac{d^2y}{dx^2}\) at \(\theta = \frac{\pi}{4}\).

---

### Solution:

We will follow these steps to solve the problem:

#### 1. **Find the first derivative, \(\frac{dy}{dx}\)**:

Using the chain rule, we first find \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) and then divide them.

##### \(\frac{dx}{d\theta}\):
\[
x = 5 + \sec 3\theta \Rightarrow \frac{dx}{d\theta} = 3\sec 3\theta \tan 3\theta.
\]

##### \(\frac{dy}{d\theta}\):
\[
y = 1 - \tan 3\theta \Rightarrow \frac{dy}{d\theta} = -3\sec^2 3\theta.
\]

Thus, the first derivative \(\frac{dy}{dx}\) is:
\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-3\sec^2 3\theta}{3\sec 3\theta \tan 3\theta} = \frac{-\sec 3\theta}{\tan 3\theta} = -\cot 3\theta.
\]

#### 2. **Find the second derivative, \(\frac{d^2y}{dx^2}\)**:

To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(\theta\), and then divide by \(\frac{dx}{d\theta}\).

##### Differentiate \(\frac{dy}{dx} = -\cot 3\theta\):
\[
\frac{d}{d\theta}(-\cot 3\theta) = 3\csc^2 3\theta.
\]

Now, divide by \(\frac{dx}{d\theta} = 3\sec 3\theta \tan 3\theta\):
\[
\frac{d^2y}{dx^2} = \frac{3\csc^2 3\theta}{3\sec 3\theta \tan 3\theta} = \frac{\csc^2 3\theta}{\sec 3\theta \tan 3\theta}.
\]

#### 3. **Evaluate at \(\theta = \frac{\pi}{4}\)**:

At \(\theta = \frac{\pi}{4}\), we have:
\[
3\theta = \frac{3\pi}{4}, \quad \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2}, \quad \tan\left(\frac{3\pi}{4}\right) = -1, \quad \csc^2\left(\frac{3\pi}{4}\right) = 2.
\]

Thus, at \(\theta = \frac{\pi}{4}\):
\[
\frac{d^2y}{dx^2} = \frac{2}{(-\sqrt{2})(-1)} = \frac{2}{\sqrt{2}} = \sqrt{2}.
\]

---

### Final Answer:
\[
\frac{d^2y}{dx^2} \text{ at } \theta = \frac{\pi}{4} = \sqrt{2}.
\]

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Would you like further details or have any other questions?

Here are some related questions to explore further:
1. What is the geometric meaning of the second derivative \(\frac{d^2y}{dx^2}\)?
2. How does the chain rule apply when functions are dependent on a third variable?
3. What are the trigonometric identities for \(\sec\), \(\csc\), and \(\cot\) at different angles?
4. What is the significance of evaluating derivatives at specific points like \(\theta = \frac{\pi}{4}\)?
5. How does \(\frac{dy}{dx}\) represent the rate of change in parametric equations?

**Tip**: When differentiating trigonometric functions, always double-check the identities and signs of the functions.

If x = 5 + Sec(3θ) and y = 1 - Tan(3θ), then find d²y/dx² at θ = π/4.

Learn how to calculate the second derivative d²y/dx² for a parametric equation where x = 5 + Sec(3θ) and y = 1 - Tan(3θ). This problem covers concepts in trigonometry and calculus, specifically using the chain rule and evaluating derivatives at θ = π/4. Suitable for advanced high school students in Grades 11-12.

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