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

Explanation

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

### Step 1: Rewrite sec and tan in terms of sin and cos
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting these into the equation gives:
\[ \sqrt{2} \cdot \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = 1 \]

### Step 2: Combine the fractions
Since both terms on the left side have the same denominator, we can combine them:
\[ \frac{\sqrt{2} + \sin \theta}{\cos \theta} = 1 \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ \sqrt{2} + \sin \theta = \cos \theta \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:
\[ \cos \theta - \sin \theta = \sqrt{2} \]

### Step 5: Divide the entire equation by \(\sqrt{2}\)
Dividing through by \(\sqrt{2}\) gives:
\[ \frac{\cos \theta}{\sqrt{2}} - \frac{\sin \theta}{\sqrt{2}} = 1 \]

### Step 6: Recognize the trigonometric identity
We know that:
\[ \frac{1}{\sqrt{2}} = \cos \frac{\pi}{4} \quad \text{and} \quad \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \]
Thus, we can rewrite the left side using the cosine of a sum identity:
\[ \cos \frac{\pi}{4} \cos \theta - \sin \frac{\pi}{4} \sin \theta = 1 \]
This simplifies to:
\[ \cos\left(\theta + \frac{\pi}{4}\right) = 1 \]

### Step 7: Solve for \(\theta\)
The general solution for \( \cos x = 1 \) is:
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
\[ \theta + \frac{\pi}{4} = 2n\pi \]
Rearranging gives:
\[ \theta = 2n\pi - \frac{\pi}{4} \]

### Final Answer
The solution to the equation \( \sqrt{2} \sec \theta + \tan \theta = 1 \) is:
\[ \theta = 2n\pi - \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]