If `tanalpha,tanbeta` are the roots of `x^(2)+px+q=0(pneq)` then `tan(alpha+beta)=?`

Options

A. `p-1`
B. `(p)/(q-1)`
C. `2q-p`
D. none of these

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 \(x^2 + px + q = 0\).

### Step-by-step Solution:

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

2. **Use Vieta's Formulas**:
According to Vieta's formulas:
- The sum of the roots (\(\tan \alpha + \tan \beta\)) is given by:
\[ \tan \alpha + \tan \beta = -\frac{p}{1} \]
- The product of the roots (\(\tan \alpha \cdot \tan \beta\)) is given by:
\[ \tan \alpha \cdot \tan \beta = \frac{q}{1} \]

3. **Apply the Tangent Addition Formula**:
We can use the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

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

5. **Final Expression**:
Therefore, the value of \(\tan(\alpha + \beta)\) is:
\[ \tan(\alpha + \beta) = \frac{-p}{1 - q} \]

### Conclusion:
The final answer is:
\[ \tan(\alpha + \beta) = \frac{-p}{1 - q} \]