If `x=a cos^3 theta` and `y=a sin^3 theta`, then `1+(dy/dx)^2` is equal to:

Options

A. `tan theta`
B. `tan^2 theta`
C. 1
D. `sec^2 theta`

Answer

To solve the problem, we need to find the expression for \(1 + \left(\frac{dy}{dx}\right)^2\) given that \(x = a \cos^3 \theta\) and \(y = a \sin^3 \theta\).

### Step-by-Step Solution:

1. **Express \(x\) and \(y\) in terms of \(\theta\)**:
\[ x = a \cos^3 \theta \quad \text{and} \quad y = a \sin^3 \theta \]

2. **Differentiate \(x\) and \(y\) with respect to \(\theta\)**:
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

3. **Use the chain rule to find \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
Simplifying this gives:
\[ \frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

4. **Calculate \(1 + \left(\frac{dy}{dx}\right)^2\)**:
\[ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + (-\tan \theta)^2 = 1 + \tan^2 \theta \]

5. **Use the trigonometric identity**:
We know from trigonometric identities that:
\[ 1 + \tan^2 \theta = \sec^2 \theta \]

6. **Final Result**:
Therefore, we conclude that:
\[ 1 + \left(\frac{dy}{dx}\right)^2 = \sec^2 \theta \]

### Answer:
\[ 1 + \left(\frac{dy}{dx}\right)^2 = \sec^2 \theta \]