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 Trigonometric Functions
We start by rewriting the secant and tangent functions in terms of sine and cosine:
\[ \sec \theta = \frac{1}{\cos \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting these into the equation gives:
\[ \frac{1}{\cos \theta} - 1 = (\sqrt{2} - 1) \frac{\sin \theta}{\cos \theta} \]

### Step 2: Clear the Denominator
Multiply both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ 1 - \cos \theta = (\sqrt{2} - 1) \sin \theta \]

### Step 3: Rearrange the Equation
Rearranging gives:
\[ 1 - \cos \theta - (\sqrt{2} - 1) \sin \theta = 0 \]

### Step 4: Use the Identity for Cosine
Using the identity \( \cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} \), we can express \( 1 - \cos \theta \) as:
\[ 1 - (1 - 2 \sin^2 \frac{\theta}{2}) = 2 \sin^2 \frac{\theta}{2} \]
Substituting this into the equation gives:
\[ 2 \sin^2 \frac{\theta}{2} - (\sqrt{2} - 1) \sin \theta = 0 \]

### Step 5: Substitute for Sine
Using \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \), we rewrite the equation:
\[ 2 \sin^2 \frac{\theta}{2} - (\sqrt{2} - 1)(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) = 0 \]
Factoring out \( 2 \sin \frac{\theta}{2} \):
\[ 2 \sin \frac{\theta}{2} \left( \sin \frac{\theta}{2} - (\sqrt{2} - 1) \cos \frac{\theta}{2} \right) = 0 \]

### Step 6: Solve the Factors
This gives us two cases to solve:
1. \( \sin \frac{\theta}{2} = 0 \)
2. \( \sin \frac{\theta}{2} - (\sqrt{2} - 1) \cos \frac{\theta}{2} = 0 \)

### Step 7: Solve Case 1
For \( \sin \frac{\theta}{2} = 0 \):
\[ \frac{\theta}{2} = n\pi \quad \Rightarrow \quad \theta = 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 8: Solve Case 2
For \( \sin \frac{\theta}{2} = (\sqrt{2} - 1) \cos \frac{\theta}{2} \):
Dividing both sides by \( \cos \frac{\theta}{2} \) (assuming \( \cos \frac{\theta}{2} \neq 0 \)):
\[ \tan \frac{\theta}{2} = \sqrt{2} - 1 \]
The angle whose tangent is \( \sqrt{2} - 1 \) is \( \frac{\pi}{8} \):
\[ \frac{\theta}{2} = \frac{\pi}{8} + n\pi \quad \Rightarrow \quad \theta = \frac{\pi}{4} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Final Solutions
Combining both cases, the solutions for \( \theta \) are:
1. \( \theta = 2n\pi \)
2. \( \theta = \frac{\pi}{4} + 2n\pi \)