Find the value of tan `pi/8`.

Explanation

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

### Step 1: Use the double angle identity
We know that:
\[ \tan(2x) = \frac{2\tan x}{1 - \tan^2 x} \]
We can set \( x = \frac{\pi}{8} \). Therefore, \( 2x = \frac{\pi}{4} \).

### Step 2: Substitute into the identity
Substituting \( x = \frac{\pi}{8} \) into the identity, we have:
\[ \tan\left(\frac{\pi}{4}\right) = \frac{2\tan\left(\frac{\pi}{8}\right)}{1 - \tan^2\left(\frac{\pi}{8}\right)} \]

### Step 3: Know the value of \( \tan\left(\frac{\pi}{4}\right) \)
We know that:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
So we can write:
\[ 1 = \frac{2\tan\left(\frac{\pi}{8}\right)}{1 - \tan^2\left(\frac{\pi}{8}\right)} \]

### Step 4: Let \( y = \tan\left(\frac{\pi}{8}\right) \)
Let \( y = \tan\left(\frac{\pi}{8}\right) \). Then we can rewrite the equation as:
\[ 1 = \frac{2y}{1 - y^2} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 1 - y^2 = 2y \]

### Step 6: Rearrange the equation
Rearranging the equation leads to:
\[ y^2 + 2y - 1 = 0 \]

### Step 7: Solve the quadratic equation
Now we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 2, c = -1 \):
\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ y = \frac{-2 \pm \sqrt{4 + 4}}{2} \]
\[ y = \frac{-2 \pm \sqrt{8}}{2} \]
\[ y = \frac{-2 \pm 2\sqrt{2}}{2} \]
\[ y = -1 \pm \sqrt{2} \]

### Step 8: Determine the correct value
This gives us two potential solutions:
\[ y = -1 + \sqrt{2} \quad \text{and} \quad y = -1 - \sqrt{2} \]
Since \( \tan\left(\frac{\pi}{8}\right) \) must be positive (as \( \frac{\pi}{8} \) is in the first quadrant), we take:
\[ \tan\left(\frac{\pi}{8}\right) = -1 + \sqrt{2} \]

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