What is the principal value of `tan^(-1)("tan"(9pi)/8)`?

Options

A. `pi/8`
B. `pi/4`
C. `pi/2`
D. `pi/6`

Answer

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

### Step 1: Identify the angle
We start with the expression \( \tan(9\pi/8) \). The angle \( 9\pi/8 \) is greater than \( \pi \) (or \( 8\pi/8 \)), which means it is in the third quadrant.

### Step 2: Use the periodic property of tangent
The tangent function has a periodicity of \( \pi \). Therefore, we can reduce the angle \( 9\pi/8 \) by subtracting \( \pi \):
\[ 9\pi/8 - \pi = 9\pi/8 - 8\pi/8 = \pi/8 \]
Thus, we have:
\[ \tan(9\pi/8) = \tan(\pi/8) \]

### Step 3: Apply the inverse tangent function
Now we can substitute back into our original expression:
\[ \tan^{-1}(\tan(9\pi/8)) = \tan^{-1}(\tan(\pi/8)) \]

### Step 4: Find the principal value
The principal value of \( \tan^{-1}(\tan(\theta)) \) is \( \theta \) if \( \theta \) is in the range \( (-\pi/2, \pi/2) \). Since \( \pi/8 \) is within this range, we have:
\[ \tan^{-1}(\tan(\pi/8)) = \pi/8 \]

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