If `x=atan theta` and `y=bsec theta`, then

Options

A. `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1`
B. `(x^(2))/(a^(2))+(y^(2))/(b^(2)=1`
C. `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
D. `(x^(2))/(a^(2))-(y^(2))/(b^(2))=0`

Answer

To solve the problem where \( x = a \tan \theta \) and \( y = b \sec \theta \), we will derive the relationship between \( x \) and \( y \) step by step.

### Step-by-Step Solution:

1. **Start with the given equations**:
\[ x = a \tan \theta \quad \text{and} \quad y = b \sec \theta \]

2. **Rewrite the equations in terms of \(\tan \theta\) and \(\sec \theta\)**:
\[ \frac{x}{a} = \tan \theta \quad \text{and} \quad \frac{y}{b} = \sec \theta \]

3. **Square both sides of the equations**:
\[ \left(\frac{x}{a}\right)^2 = \tan^2 \theta \quad \text{and} \quad \left(\frac{y}{b}\right)^2 = \sec^2 \theta \]

4. **Express \(\tan^2 \theta\) and \(\sec^2 \theta\)**:
\[ \frac{x^2}{a^2} = \tan^2 \theta \quad \text{and} \quad \frac{y^2}{b^2} = \sec^2 \theta \]

5. **Use the identity \(\sec^2 \theta = 1 + \tan^2 \theta\)**:
\[ \sec^2 \theta - 1 = \tan^2 \theta \]

6. **Substituting this identity into the equations**:
\[ \frac{x^2}{a^2} = \sec^2 \theta - 1 \]

7. **Now we can express the difference**:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = \tan^2 \theta - \sec^2 \theta \]

8. **Substituting the identity back**:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = \tan^2 \theta - (1 + \tan^2 \theta) \]
\[ = -1 \]

9. **Rearranging the equation**:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \implies \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]

### Final Result:
Thus, we conclude that:
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]