Write the principal value of `"tan"^(-1)("tan"(9pi)/(8))`.

Answer

To find the principal value of \( \tan^{-1}(\tan(9\pi/8)) \), we will follow these steps:

### Step 1: Identify the angle
We start with the angle \( \frac{9\pi}{8} \).

### Step 2: Check the range
The principal value of \( \tan^{-1}(x) \) is defined for \( x \) in the range of \( -\frac{\pi}{2} < y < \frac{\pi}{2} \). Since \( \frac{9\pi}{8} \) is greater than \( \frac{\pi}{2} \), we need to adjust this angle.

### Step 3: Use the periodic property of tangent
We can express \( \frac{9\pi}{8} \) in a different form:
\[ \frac{9\pi}{8} = \pi + \frac{\pi}{8} \]
Using the property of the tangent function, we know:
\[ \tan(\pi + \theta) = \tan(\theta) \]
Thus,
\[ \tan\left(\frac{9\pi}{8}\right) = \tan\left(\pi + \frac{\pi}{8}\right) = \tan\left(\frac{\pi}{8}\right) \]

### Step 4: Substitute back into the inverse tangent
Now we can rewrite our original expression:
\[ \tan^{-1}\left(\tan\left(\frac{9\pi}{8}\right)\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{8}\right)\right) \]

### Step 5: Find the principal value
Since \( \frac{\pi}{8} \) is within the range \( -\frac{\pi}{2} < \frac{\pi}{8} < \frac{\pi}{2} \), we can directly state:
\[ \tan^{-1}\left(\tan\left(\frac{\pi}{8}\right)\right) = \frac{\pi}{8} \]

### Final Answer
Thus, the principal value of \( \tan^{-1}(\tan(9\pi/8)) \) is:
\[ \frac{\pi}{8} \]

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