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 and their properties**:
The roots of the quadratic equation are given as \(\sin \theta\) and \(\cos \theta\).

2. **Use the sum and product of roots**:
From the properties of quadratic equations, we know:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
- The product of the roots \(\alpha \beta = \frac{c}{a}\)

Here, we have:
\[ \sin \theta + \cos \theta = -\frac{b}{a} \]
\[ \sin \theta \cdot \cos \theta = \frac{c}{a} \]

3. **Square the sum of the roots**:
Now, 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 Pythagorean identity, \(\sin^2 \theta + \cos^2 \theta = 1\), we can substitute:
\[ 1 + 2 \sin \theta \cos \theta = \left(-\frac{b}{a}\right)^2 \]
Thus,
\[ 1 + 2 \cdot \frac{c}{a} = \frac{b^2}{a^2} \]

4. **Rearranging the equation**:
Multiply through by \(a^2\) to eliminate the denominator:
\[ a^2 + 2ac = b^2 \]

5. **Express \((a+c)^2\)**:
We can express \((a+c)^2\) as:
\[ (a+c)^2 = a^2 + 2ac + c^2 \]
From the previous step, we know \(a^2 + 2ac = b^2\), so we can substitute:
\[ (a+c)^2 = b^2 + c^2 \]

6. **Substituting into the original expression**:
Now, we substitute this into the expression we need to evaluate:
\[ \frac{(a+c)^2}{b^2+c^2} = \frac{b^2 + c^2}{b^2 + c^2} = 1 \]

### Final Answer:
Thus, the value of \(\frac{(a+c)^2}{b^2+c^2}\) is \(\boxed{1}\).