If `alpha and beta (alpha gt beta)` are the roots of `x^(2) + kx - 1 =0`, then find the value of `tan^(-1) alpha - tan^(-1) beta`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots \( \alpha \) and \( \beta \) of the quadratic equation \( x^2 + kx - 1 = 0 \) can be found using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = k \), and \( c = -1 \). Thus, the roots are:
\[ \alpha, \beta = \frac{-k \pm \sqrt{k^2 + 4}}{2} \]

2. **Calculate \( \alpha \beta \)**:
From Vieta's formulas, we know:
\[ \alpha \beta = c/a = -1 \]

3. **Use the Formula for \( \tan^{-1} \)**:
We can use the identity for the difference of inverse tangents:
\[ \tan^{-1} \alpha - \tan^{-1} \beta = \tan^{-1} \left( \frac{\alpha - \beta}{1 + \alpha \beta} \right) \]

4. **Substitute \( \alpha \beta \)**:
Since \( \alpha \beta = -1 \), we can substitute this into the formula:
\[ \tan^{-1} \alpha - \tan^{-1} \beta = \tan^{-1} \left( \frac{\alpha - \beta}{1 - 1} \right) \]
This leads to a division by zero, which indicates that the expression tends towards infinity.

5. **Conclusion**:
Since \( \tan^{-1} \left( \frac{\alpha - \beta}{0} \right) \) approaches infinity, we conclude:
\[ \tan^{-1} \alpha - \tan^{-1} \beta = \frac{\pi}{2} \]

### Final Answer:
\[ \tan^{-1} \alpha - \tan^{-1} \beta = \frac{\pi}{2} \]