If tan A and tan B are the roots of the quadratic equation `x^(2)-px+q=0`, then the value `sin^(2)(A+B)` is

Options

A. `(p^(2))/(p^(2)+q^(2))`
B. `(p^(2))/((p+q)^(2))`
C. `1-(p)/((1-q)^(2))`
D. `(p^(2))/((1-q)^(2)+p^(2))`

Answer

To solve the problem, we need to find the value of \( \sin^2(A+B) \) given that \( \tan A \) and \( \tan B \) are the roots of the quadratic equation \( x^2 - px + q = 0 \).

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the quadratic equation \( x^2 - px + q = 0 \) are \( \tan A \) and \( \tan B \).

2. **Sum and Product of Roots**:
From the properties of quadratic equations, we know:
- The sum of the roots \( \tan A + \tan B = p \)
- The product of the roots \( \tan A \tan B = q \)

3. **Use the Formula for \( \tan(A+B) \)**:
We can use the formula for the tangent of the sum of two angles:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values we found:
\[ \tan(A+B) = \frac{p}{1 - q} \]

4. **Find \( \sin(A+B) \)**:
We know that:
\[ \sin(A+B) = \frac{\tan(A+B)}{\sqrt{1 + \tan^2(A+B)}} \]
Therefore, we need to find \( \tan^2(A+B) \):
\[ \tan^2(A+B) = \left(\frac{p}{1 - q}\right)^2 \]
Now, substituting this into the equation for \( \sin(A+B) \):
\[ \sin(A+B) = \frac{\frac{p}{1 - q}}{\sqrt{1 + \left(\frac{p}{1 - q}\right)^2}} \]

5. **Simplify the Expression**:
The expression under the square root simplifies to:
\[ 1 + \left(\frac{p}{1 - q}\right)^2 = \frac{(1 - q)^2 + p^2}{(1 - q)^2} \]
Thus,
\[ \sin(A+B) = \frac{p}{1 - q} \cdot \frac{(1 - q)}{\sqrt{(1 - q)^2 + p^2}} = \frac{p}{\sqrt{(1 - q)^2 + p^2}} \]

6. **Find \( \sin^2(A+B) \)**:
Now, squaring \( \sin(A+B) \):
\[ \sin^2(A+B) = \left(\frac{p}{\sqrt{(1 - q)^2 + p^2}}\right)^2 = \frac{p^2}{(1 - q)^2 + p^2} \]

### Final Result:
Thus, the value of \( \sin^2(A+B) \) is:
\[ \sin^2(A+B) = \frac{p^2}{(1 - q)^2 + p^2} \]