If `sin theta, cos theta` are the roots of the equation `ax^(2)+bx+c=0` then find the value of `((a+c)^(2))/(b^(2)+c^(2))`

Answer

To solve the problem, we need to find the value of \(\frac{(a+c)^2}{b^2 + c^2}\) given that \(\sin \theta\) and \(\cos \theta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step-by-Step Solution:

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

2. **Use the Sum and Product of Roots**:
According to Vieta's formulas:
- The sum of the roots \(\sin \theta + \cos \theta = -\frac{b}{a}\)
- The product of the roots \(\sin \theta \cdot \cos \theta = \frac{c}{a}\)

3. **Square the Sum of Roots**:
We square the sum of the roots:
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ 1 + 2\sin \theta \cos \theta = \left(-\frac{b}{a}\right)^2 \]

4. **Substitute the Product of Roots**:
From the product of the roots, we know:
\[ 2\sin \theta \cos \theta = 2 \cdot \frac{c}{a} = \frac{2c}{a} \]
Substituting this into the equation gives:
\[ 1 + \frac{2c}{a} = \frac{b^2}{a^2} \]

5. **Rearranging the Equation**:
Multiply through by \(a^2\) to eliminate the denominators:
\[ a^2 + 2ac = b^2 \]

6. **Add \(c^2\) to Both Sides**:
Now, we add \(c^2\) to both sides:
\[ a^2 + 2ac + c^2 = b^2 + c^2 \]
The left-hand side can be factored:
\[ (a+c)^2 = b^2 + c^2 \]

7. **Final Expression**:
Therefore, we can express \(\frac{(a+c)^2}{b^2 + c^2}\):
\[ \frac{(a+c)^2}{b^2 + c^2} = \frac{b^2 + c^2}{b^2 + c^2} = 1 \]

### Conclusion:
The value of \(\frac{(a+c)^2}{b^2 + c^2}\) is \(1\).