If `"sec"^(2) theta = sqrt(2) (1-"tan"^(2) theta), "then" theta=`

Options

A. `n pi + (pi)/(8), n in Z`
B. `n pi +-(pi)/(4), n in Z`
C. `n pi +-(pi)/(8), n in Z`
D. none of these

Answer

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

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} \quad \text{and} \quad \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Substituting these identities into the equation gives:
\[ \frac{1}{\cos^2 \theta} = \sqrt{2} \left( 1 - \frac{\sin^2 \theta}{\cos^2 \theta} \right) \]

### Step 2: Simplify the right-hand side
To simplify the right-hand side, we can express it with a common denominator:
\[ \frac{1}{\cos^2 \theta} = \sqrt{2} \left( \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta} \right) \]
This simplifies to:
\[ \frac{1}{\cos^2 \theta} = \frac{\sqrt{2} (\cos^2 \theta - \sin^2 \theta)}{\cos^2 \theta} \]

### Step 3: Cancel the common terms
Since both sides have the term \( \frac{1}{\cos^2 \theta} \), we can cancel \( \cos^2 \theta \) from both sides (assuming \( \cos^2 \theta \neq 0 \)):
\[ 1 = \sqrt{2} (\cos^2 \theta - \sin^2 \theta) \]

### Step 4: Use the double angle identity
Recall the double angle identity:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \]
Thus, we can rewrite the equation as:
\[ 1 = \sqrt{2} \cos 2\theta \]

### Step 5: Solve for \( \cos 2\theta \)
Rearranging gives:
\[ \cos 2\theta = \frac{1}{\sqrt{2}} \]

### Step 6: Find the general solution for \( 2\theta \)
The cosine function equals \( \frac{1}{\sqrt{2}} \) at:
\[ 2\theta = \frac{\pi}{4} + 2n\pi \quad \text{and} \quad 2\theta = -\frac{\pi}{4} + 2n\pi \]
where \( n \) is any integer.

### Step 7: Divide by 2 to find \( \theta \)
Dividing the entire equation by 2 gives:
\[ \theta = \frac{\pi}{8} + n\pi \quad \text{and} \quad \theta = -\frac{\pi}{8} + n\pi \]

### Final Solution
Thus, the general solutions for \( \theta \) are:
\[ \theta = n\pi + \frac{\pi}{8} \quad \text{and} \quad \theta = n\pi - \frac{\pi}{8} \]