To find the value of $\cot (\tan^{-1} \alpha + \cot^{-1} \alpha)$, we start by defining some variables for clarity.

Let: $\theta = \tan^{-1} \alpha$ and $\phi = \cot^{-1} \alpha$

Thus: $\alpha = \tan \theta$ and $\alpha = \cot \phi$

Since $\cot \phi = \frac{1}{\tan \phi}$, we have: $\alpha = \frac{1}{\tan \phi} \implies \tan \phi = \frac{1}{\alpha}$

Next, we need to find $\theta + \phi$. To do this, we use the fact that $\tan \theta = \alpha$ and $\tan \phi = \frac{1}{\alpha}$. Using the angle addition formula for tangent, we get: $\tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}$

Substituting $\tan \theta = \alpha$ and $\tan \phi = \frac{1}{\alpha}$: $\tan(\theta + \phi) = \frac{\alpha + \frac{1}{\alpha}}{1 - \alpha \cdot \frac{1}{\alpha}} = \frac{\alpha + \frac{1}{\alpha}}{1 - 1}$

Since the denominator is zero, $\tan(\theta + \phi)$ is undefined, which implies that $\theta + \phi = \frac{\pi}{2}$ because $\tan \frac{\pi}{2}$ is undefined.

Now, we want to find $\cot (\theta + \phi)$: $\cot (\theta + \phi) = \cot \left( \frac{\pi}{2} \right)$

Since $\cot \frac{\pi}{2} = 0$: $\cot (\tan^{-1} \alpha + \cot^{-1} \alpha) = 0$

Thus, the value is: $\boxed{0}$

Would you like details on any part of this solution, or do you have any other questions? Here are some related questions you might find interesting:

1. Find the value of $\sin (\tan^{-1} x + \cot^{-1} x)$.
2. Simplify $\tan (\sin^{-1} x + \cos^{-1} x)$.
3. Solve for $\alpha$ in $\cot (\sin^{-1} \alpha + \cos^{-1} \alpha) = 1$.
4. Evaluate $\cot (\sec^{-1} x + \csc^{-1} x)$.
5. Find the value of $\tan (\cos^{-1} x + \sin^{-1} x)$.

Tip: When working with inverse trigonometric functions, remember their domains and ranges to avoid undefined expressions or incorrect simplifications.