If `alpha in (-(pi)/(2), 0)`, then find the value of `tan^(-1) (cot alpha) - cot^(-1) (tan alpha)`

Explanation

To solve the problem, we need to find the value of \( \tan^{-1}(\cot \alpha) - \cot^{-1}(\tan \alpha) \) given that \( \alpha \in \left(-\frac{\pi}{2}, 0\right) \).

### Step-by-Step Solution:

1. **Express cotangent in terms of tangent:**
\[ \cot \alpha = \frac{1}{\tan \alpha} \]
Therefore, we can rewrite the expression as:
\[ \tan^{-1}(\cot \alpha) = \tan^{-1}\left(\frac{1}{\tan \alpha}\right) \]

2. **Use the identity for the inverse tangent:**
We know that:
\[ \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} - \tan^{-1}(x) \quad \text{for } x > 0 \]
However, since \( \tan \alpha < 0 \) in the interval \( \left(-\frac{\pi}{2}, 0\right) \), we need to adjust this identity:
\[ \tan^{-1}\left(\frac{1}{\tan \alpha}\right) = -\frac{\pi}{2} + \cot^{-1}(\tan \alpha) \]

3. **Substituting back into the expression:**
Now substituting this back into our original expression:
\[ \tan^{-1}(\cot \alpha) - \cot^{-1}(\tan \alpha) = \left(-\frac{\pi}{2} + \cot^{-1}(\tan \alpha)\right) - \cot^{-1}(\tan \alpha) \]

4. **Simplifying the expression:**
The \( \cot^{-1}(\tan \alpha) \) terms cancel out:
\[ = -\frac{\pi}{2} \]

5. **Final Result:**
Therefore, the value of \( \tan^{-1}(\cot \alpha) - \cot^{-1}(\tan \alpha) \) is:
\[ \boxed{-\frac{\pi}{2}} \]