If `(x)/(a)costheta+(y)/(b)sintheta=1&(x)/(a)sintheta-(y)/(b)costheta=1`. Then

Options

A. `(x^(2))/(a^(2))+(y^(2))/(b^(2))=0`
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))=2`

Answer

To solve the given equations:

1. **Write down the equations:**
\[ \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \tag{1} \]
\[ \frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta = 1 \tag{2} \]

2. **Identify coefficients:**
From equation (1), the coefficients are:
- \( A = \frac{x}{a} \) (coefficient of \(\cos \theta\))
- \( B = \frac{y}{b} \) (coefficient of \(\sin \theta\))

From equation (2), the coefficients are:
- \( C = \frac{x}{a} \) (coefficient of \(\sin \theta\))
- \( D = -\frac{y}{b} \) (coefficient of \(-\cos \theta\))

3. **Recognize the pattern:**
The equations can be rewritten in a form that suggests a relationship between the coefficients:
\[ A \cos \theta + B \sin \theta = 1 \]
\[ C \sin \theta + D \cos \theta = 1 \]

Notice that \( A \) and \( D \) are related to \( C \) and \( B \) in a specific way.

4. **Apply the identity:**
From the given structure of the equations, we can use the identity:
\[ A^2 + B^2 = C^2 + D^2 \]
where \( A = \frac{x}{a} \), \( B = \frac{y}{b} \), \( C = \frac{x}{a} \), and \( D = -\frac{y}{b} \).

5. **Substitute the coefficients:**
Substitute the coefficients into the identity:
\[ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = \left(\frac{x}{a}\right)^2 + \left(-\frac{y}{b}\right)^2 \]

6. **Simplify the equation:**
This simplifies to:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{x^2}{a^2} + \frac{y^2}{b^2} \]
which is trivially true.

7. **Final result:**
The sum of the squares of the coefficients gives us:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \]

Thus, the final result is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \]