Evaluate the following
`tan^(-1) { tan (- (7pi)/8)}`.

Answer

To evaluate \( \tan^{-1} \left( \tan \left( -\frac{7\pi}{8} \right) \right) \), we will follow these steps:

### Step 1: Use the property of tangent
We know that:
\[ \tan(-\theta) = -\tan(\theta) \]
Thus, we can rewrite:
\[ \tan \left( -\frac{7\pi}{8} \right) = -\tan \left( \frac{7\pi}{8} \right) \]
So, we have:
\[ \tan^{-1} \left( \tan \left( -\frac{7\pi}{8} \right) \right) = \tan^{-1} \left( -\tan \left( \frac{7\pi}{8} \right) \right) \]

### Step 2: Use the property of inverse tangent
We also know that:
\[ \tan^{-1}(-\theta) = -\tan^{-1}(\theta) \]
Applying this property gives us:
\[ \tan^{-1} \left( -\tan \left( \frac{7\pi}{8} \right) \right) = -\tan^{-1} \left( \tan \left( \frac{7\pi}{8} \right) \right) \]

### Step 3: Determine the value of \( \tan^{-1} \left( \tan \left( \frac{7\pi}{8} \right) \right) \)
The function \( \tan^{-1}(\tan(\theta)) \) equals \( \theta \) if \( \theta \) is in the range \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \). However, since \( \frac{7\pi}{8} \) is greater than \( \frac{\pi}{2} \), we need to adjust it.

We can express \( \frac{7\pi}{8} \) as:
\[ \frac{7\pi}{8} = \pi - \frac{\pi}{8} \]
Using the property \( \tan(\pi - \theta) = -\tan(\theta) \), we have:
\[ \tan\left(\frac{7\pi}{8}\right) = -\tan\left(\frac{\pi}{8}\right) \]

### Step 4: Substitute back into the equation
Thus, we can write:
\[ \tan^{-1} \left( \tan \left( \frac{7\pi}{8} \right) \right) = \tan^{-1} \left( -\tan \left( \frac{\pi}{8} \right) \right) \]
This leads to:
\[ \tan^{-1} \left( -\tan \left( \frac{\pi}{8} \right) \right) = -\tan^{-1} \left( \tan \left( \frac{\pi}{8} \right) \right) \]

### Step 5: Evaluate the final expression
Since \( \frac{\pi}{8} \) is within the range \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), we have:
\[ -\tan^{-1} \left( \tan \left( \frac{\pi}{8} \right) \right) = -\frac{\pi}{8} \]

### Final Result
Thus, the value of \( \tan^{-1} \left( \tan \left( -\frac{7\pi}{8} \right) \right) \) is:
\[ \boxed{-\frac{\pi}{8}} \]