Find the value of tan `pi/8`.

Explanation

To find the value of \( \tan \frac{\pi}{8} \), we can use the double angle formula for tangent. Here’s a step-by-step solution:

### Step 1: Identify the angle
We know that:
\[ \frac{\pi}{4} = 2 \cdot \frac{\pi}{8} \]
This means we can express \( \tan \frac{\pi}{4} \) in terms of \( \tan \frac{\pi}{8} \).

### Step 2: Use the double angle formula
The double angle formula for tangent is given by:
\[ \tan(2x) = \frac{2 \tan x}{1 - \tan^2 x} \]
Let \( x = \frac{\pi}{8} \). Therefore, we have:
\[ \tan \frac{\pi}{4} = \tan(2 \cdot \frac{\pi}{8}) = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \]

### Step 3: Substitute known values
We know that:
\[ \tan \frac{\pi}{4} = 1 \]
Substituting this into the equation gives:
\[ 1 = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \]

### Step 4: Let \( t = \tan \frac{\pi}{8} \)
Let’s denote \( t = \tan \frac{\pi}{8} \). The equation now becomes:
\[ 1 = \frac{2t}{1 - t^2} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 1 - t^2 = 2t \]
Rearranging this results in:
\[ t^2 + 2t - 1 = 0 \]

### Step 6: Solve the quadratic equation
Now we can use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -1 \):
\[ t = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
This simplifies to:
\[ t = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]

### Step 7: Determine the correct value
Since \( \frac{\pi}{8} \) is in the first quadrant, where tangent is positive, we take the positive root:
\[ t = -1 + \sqrt{2} \]

### Final Result
Thus, the value of \( \tan \frac{\pi}{8} \) is:
\[ \tan \frac{\pi}{8} = -1 + \sqrt{2} \]