Evaluate `cot (tan^(-1) (2x) + cot^(-1) (2x))`

Answer

To evaluate the expression \( \cot(\tan^{-1}(2x) + \cot^{-1}(2x)) \), we can follow these steps:

### Step 1: Recognize the relationship between the inverse functions
We know that:
\[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \]
for any \( x > 0 \).

### Step 2: Apply the formula
In our case, we have \( x = 2x \). Therefore, we can write:
\[ \tan^{-1}(2x) + \cot^{-1}(2x) = \frac{\pi}{2} \]

### Step 3: Substitute into the original expression
Now we substitute this result into our original expression:
\[ \cot(\tan^{-1}(2x) + \cot^{-1}(2x)) = \cot\left(\frac{\pi}{2}\right) \]

### Step 4: Evaluate \( \cot\left(\frac{\pi}{2}\right) \)
We know that:
\[ \cot\left(\frac{\pi}{2}\right) = 0 \]

### Final Answer
Thus, the value of \( \cot(\tan^{-1}(2x) + \cot^{-1}(2x)) \) is:
\[ \boxed{0} \]