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

Options

A. `tan^ 2theta`
B. `cot^ 2theta`
C. `sec^(2) theta`
D. `"cosec"^2 theta`

Answer

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

### Step-by-Step Solution:

1. **Differentiate \( x \) with respect to \( \theta \)**:
\[ x = \cos^3 \theta \]
Using the chain rule:
\[ \frac{dx}{d\theta} = 3 \cos^2 \theta (-\sin \theta) = -3 \cos^2 \theta \sin \theta \]

2. **Differentiate \( y \) with respect to \( \theta \)**:
\[ y = \sin^3 \theta \]
Again, using the chain rule:
\[ \frac{dy}{d\theta} = 3 \sin^2 \theta \cos \theta \]

3. **Find \( \frac{dy}{dx} \)**:
Using the relationship \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \):
\[ \frac{dy}{dx} = \frac{3 \sin^2 \theta \cos \theta}{-3 \cos^2 \theta \sin \theta} \]
Simplifying this expression:
\[ \frac{dy}{dx} = \frac{\sin \theta}{-\cos \theta} = -\tan \theta \]

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

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

Thus, the final result is:
\[ 1 + \left( \frac{dy}{dx} \right)^2 = \sec^2 \theta \]

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