Solve `sec theta-1=(sqrt(2)-1) tan theta`.

Explanation

To solve the equation \( \sec \theta - 1 = (\sqrt{2} - 1) \tan \theta \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore, we can rewrite the equation as:
\[ \frac{1}{\cos \theta} - 1 = (\sqrt{2} - 1) \frac{\sin \theta}{\cos \theta} \]

### Step 2: Simplify the left side
To simplify the left side, we will find a common denominator:
\[ \frac{1 - \cos \theta}{\cos \theta} = (\sqrt{2} - 1) \frac{\sin \theta}{\cos \theta} \]
Now, we can multiply both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ 1 - \cos \theta = (\sqrt{2} - 1) \sin \theta \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 1 - \cos \theta = (\sqrt{2} - 1) \sin \theta \]

### Step 4: Use the identity for sine and cosine
We can use the identity \( \sin A \sin B + \cos A \cos B = \cos(A - B) \). We can express the equation in terms of sine and cosine:
\[ \sin \theta \sin \frac{\pi}{8} + \cos \theta \cos \frac{\pi}{8} = \cos \left( \theta - \frac{\pi}{8} \right) \]
Thus, we can rewrite the equation as:
\[ \cos \left( \theta - \frac{\pi}{8} \right) = \cos \frac{\pi}{8} \]

### Step 5: Solve for \( \theta \)
The general solution for \( \cos A = \cos B \) is given by:
\[ A = 2n\pi \pm B \quad (n \in \mathbb{Z}) \]
Applying this to our equation:
\[ \theta - \frac{\pi}{8} = 2n\pi \pm \frac{\pi}{8} \]

### Step 6: Isolate \( \theta \)
From the above, we can derive two equations:
1. \( \theta - \frac{\pi}{8} = 2n\pi + \frac{\pi}{8} \)
2. \( \theta - \frac{\pi}{8} = 2n\pi - \frac{\pi}{8} \)

Solving these gives:
1. \( \theta = 2n\pi + \frac{\pi}{4} \)
2. \( \theta = 2n\pi \)

### Final Solutions
Thus, the solutions to the equation are:
\[ \theta = 2n\pi + \frac{\pi}{4} \quad \text{and} \quad \theta = 2n\pi \quad (n \in \mathbb{Z}) \]