Find the value of `tan^(-1)(tan'(2pi)/(3))`.

Explanation

To find the value of \( \tan^{-1}(\tan(2\pi/3)) \), we can follow these steps:

### Step 1: Evaluate \( \tan(2\pi/3) \)
The angle \( 2\pi/3 \) can be rewritten in terms of a reference angle:
\[ 2\pi/3 = \pi - \pi/3 \]
In the second quadrant, the tangent function is negative. Therefore, we have:
\[ \tan(2\pi/3) = -\tan(\pi/3) \]
Since \( \tan(\pi/3) = \sqrt{3} \), we find:
\[ \tan(2\pi/3) = -\sqrt{3} \]

### Step 2: Substitute into the inverse tangent function
Now we substitute this back into the inverse tangent function:
\[ \tan^{-1}(\tan(2\pi/3)) = \tan^{-1}(-\sqrt{3}) \]

### Step 3: Determine the angle corresponding to \( -\sqrt{3} \)
The value \( \tan^{-1}(-\sqrt{3}) \) corresponds to an angle in the fourth quadrant. The reference angle for \( \sqrt{3} \) is \( \pi/3 \), so:
\[ \tan^{-1}(-\sqrt{3}) = -\pi/3 \]

### Step 4: Final result
Thus, we conclude:
\[ \tan^{-1}(\tan(2\pi/3)) = -\frac{\pi}{3} \]

### Summary of the solution:
The final answer is:
\[ \boxed{-\frac{\pi}{3}} \]