Let `alpha, beta` are the roots of the equation `x^(2)+7x+k(k-3)=0`, where `k in (0, 3)` and k is a constant. Then the value of `tan^(-1)alpha+tan^(-1)beta+"tan"^(-1)(1)/(alpha)+"tan"^(-1)(1)/(beta)` is :

Options

A. `pi`
B. `(pi)/(2)`
C. 0
D. `-(pi)/(2)`

Answer

To solve the problem, we need to find the value of \( \tan^{-1} \alpha + \tan^{-1} \beta + \tan^{-1} \frac{1}{\alpha} + \tan^{-1} \frac{1}{\beta} \), where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 + 7x + k(k-3) = 0 \).

### Step 1: Identify the roots
The roots \( \alpha \) and \( \beta \) of the quadratic equation can be found using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -7 \).
- The product of the roots \( \alpha \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = k(k-3) \).

### Step 2: Use the formula for the sum of inverse tangents
We will use the formula:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \]
Applying this to \( \tan^{-1} \alpha + \tan^{-1} \beta \):
\[ \tan^{-1} \alpha + \tan^{-1} \beta = \tan^{-1} \left( \frac{\alpha + \beta}{1 - \alpha \beta} \right) = \tan^{-1} \left( \frac{-7}{1 - k(k-3)} \right) \]

### Step 3: Calculate \( \tan^{-1} \frac{1}{\alpha} + \tan^{-1} \frac{1}{\beta} \)
Using the same formula for \( \tan^{-1} \frac{1}{\alpha} + \tan^{-1} \frac{1}{\beta} \):
\[ \tan^{-1} \frac{1}{\alpha} + \tan^{-1} \frac{1}{\beta} = \tan^{-1} \left( \frac{\frac{1}{\alpha} + \frac{1}{\beta}}{1 - \frac{1}{\alpha \beta}} \right) = \tan^{-1} \left( \frac{\frac{\beta + \alpha}{\alpha \beta}}{1 - \frac{1}{\alpha \beta}} \right) \]
Substituting \( \alpha + \beta = -7 \) and \( \alpha \beta = k(k-3) \):
\[ \tan^{-1} \left( \frac{-7/k(k-3)}{1 - \frac{1}{k(k-3)}} \right) \]
This simplifies to:
\[ \tan^{-1} \left( \frac{-7}{k(k-3) - 1} \right) \]

### Step 4: Combine the results
Now we combine both results:
\[ \tan^{-1} \left( \frac{-7}{1 - k(k-3)} \right) + \tan^{-1} \left( \frac{-7}{k(k-3) - 1} \right) \]
Using the property \( \tan^{-1} x + \tan^{-1} (-x) = 0 \):
\[ \tan^{-1} \left( \frac{-7}{1 - k(k-3)} \right) + \tan^{-1} \left( \frac{7}{1 - k(k-3)} \right) = 0 \]

### Final Result
Thus, the value of \( \tan^{-1} \alpha + \tan^{-1} \beta + \tan^{-1} \frac{1}{\alpha} + \tan^{-1} \frac{1}{\beta} \) is:
\[ \boxed{0} \]