If `sintheta` and `costheta` are roots of equation `ax^(2) + bx + c =0`, then

Options

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

Answer

To solve the problem, we need to analyze the relationship between the roots of the quadratic equation \( ax^2 + bx + c = 0 \) and the given roots \( \sin \theta \) and \( \cos \theta \).

### Step-by-Step Solution:

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

2. **Use the Relationship Between Roots and Coefficients**:
For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the sum of the roots \( r_1 + r_2 \) is given by:
\[ r_1 + r_2 = -\frac{b}{a} \]
Therefore, we can write:
\[ \sin \theta + \cos \theta = -\frac{b}{a} \tag{1} \]

3. **Square the Sum of the Roots**:
Squaring both sides of equation (1):
\[ (\sin \theta + \cos \theta)^2 = \left(-\frac{b}{a}\right)^2 \]
Expanding the left-hand side:
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{b^2}{a^2} \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2\sin \theta \cos \theta = \frac{b^2}{a^2} \tag{2} \]

4. **Use the Product of the Roots**:
The product of the roots \( r_1 \cdot r_2 \) is given by:
\[ r_1 \cdot r_2 = \frac{c}{a} \]
Thus, we have:
\[ \sin \theta \cos \theta = \frac{c}{a} \tag{3} \]

5. **Substitute the Product into Equation (2)**:
From equation (3), we can substitute \( \sin \theta \cos \theta \) into equation (2):
\[ 1 + 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} \]
Simplifying this gives:
\[ 1 + \frac{2c}{a} = \frac{b^2}{a^2} \]

6. **Multiply Through by \( a^2 \)**:
To eliminate the fractions, multiply through by \( a^2 \):
\[ a^2 + 2ac = b^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ a^2 + 2ac - b^2 = 0 \]
Adding \( c^2 \) to both sides:
\[ a^2 + 2ac + c^2 = b^2 + c^2 \]

8. **Final Form**:
The left-hand side can be factored:
\[ (a + c)^2 = b^2 + c^2 \]

Thus, we conclude that:
\[ a + c^2 = b^2 + c^2 \]

### Conclusion:
The correct relationship derived from the roots \( \sin \theta \) and \( \cos \theta \) is:
\[ a + c^2 = b^2 + c^2 \]