If `sintheta and costheta` are the roots of the equation`ax^(2)-bx+c=0` , then which of the following selection is correct :

Options

A. `a^(2)+b^(2)=2ac=0`
B. `a^(2)-b^(2)+2ac=0`
C. `a^(2)+b^(2)-2ac=0`
D. `a^(2)-b^(2)-2ac=0`

Answer

To solve the problem where \( \sin \theta \) and \( \cos \theta \) are the roots of the quadratic equation \( ax^2 - bx + c = 0 \), we will use Vieta's formulas, which relate the coefficients of the polynomial to sums and products of its roots.

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the equation are given as \( \sin \theta \) and \( \cos \theta \).

2. **Apply Vieta's Formulas**:
According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = \sin \theta + \cos \theta \) is equal to \( -\frac{b}{a} \).
- The product of the roots \( \alpha \cdot \beta = \sin \theta \cdot \cos \theta \) is equal to \( \frac{c}{a} \).

3. **Write the Equations**:
From the above, we can write:
\[ \sin \theta + \cos \theta = -\frac{b}{a} \quad (1) \]
\[ \sin \theta \cdot \cos \theta = \frac{c}{a} \quad (2) \]

4. **Express \( \sin \theta \cdot \cos \theta \)**:
We know that \( \sin \theta \cdot \cos \theta = \frac{1}{2} \sin(2\theta) \). Thus, we can also express equation (2) as:
\[ \frac{1}{2} \sin(2\theta) = \frac{c}{a} \quad (3) \]

5. **Square the Sum of the Roots**:
We can square equation (1):
\[ (\sin \theta + \cos \theta)^2 = \left(-\frac{b}{a}\right)^2 \]
Expanding the left side gives:
\[ \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = \frac{b^2}{a^2} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2 \sin \theta \cos \theta = \frac{b^2}{a^2} \]

6. **Substitute the Product of Roots**:
Substitute equation (2) into the squared sum:
\[ 1 + 2 \cdot \frac{c}{a} = \frac{b^2}{a^2} \]
Rearranging gives:
\[ 2c = \frac{b^2}{a^2} - 1 \quad (4) \]

7. **Final Form**:
Multiply through by \( a^2 \) to eliminate the fraction:
\[ 2ac = b^2 - a^2 \]

### Conclusion:
The relationship derived from the roots \( \sin \theta \) and \( \cos \theta \) leads us to the equation \( b^2 = 2ac + a^2 \).