Given that tan `alpha and tan beta` are the roots of the equations `x^(2) + bx + c = 0` with b not equal to 0
What is the tan `(alpha + beta)` equal to

Options

A. b/(c -1)
B. c(b -1)
C. `c(b -1)^(2)`
D. `b(c -1)^(2)`

Answer

To solve the problem, we need to find the value of \(\tan(\alpha + \beta)\) given that \(\tan \alpha\) and \(\tan \beta\) are the roots of the quadratic equation \(x^2 + bx + c = 0\).

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the equation \(x^2 + bx + c = 0\) are given as \(\tan \alpha\) and \(\tan \beta\).

2. **Use Vieta's Formulas**:
According to Vieta's formulas:
- The sum of the roots \(\tan \alpha + \tan \beta = -\frac{b}{1} = -b\).
- The product of the roots \(\tan \alpha \cdot \tan \beta = \frac{c}{1} = c\).

3. **Use the Tangent Addition Formula**:
The formula for \(\tan(\alpha + \beta)\) is given by:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

4. **Substitute the Values**:
Substitute the values we found from Vieta's formulas into the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{-b}{1 - c} \]

5. **Rearranging the Expression**:
We can also express this as:
\[ \tan(\alpha + \beta) = -\frac{b}{1 - c} = \frac{b}{c - 1} \]

6. **Final Answer**:
Therefore, the value of \(\tan(\alpha + \beta)\) is:
\[ \tan(\alpha + \beta) = \frac{b}{c - 1} \]

### Conclusion:
The correct answer is \(\frac{b}{c - 1}\).