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

Options

A. `n pi + (-1)^(n) (pi)/(4), n in Z`
B. `2n pi +-(pi)/(3), n in Z`
C. `n pi +- (2pi)/(3), n in Z`
D. `n pi- (pi)/(4), n in Z`

Answer

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

### Step 1: Rewrite the equation in terms of sine and cosine
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:
\[ \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta} = \sqrt{2} \]
This simplifies to:
\[ \frac{\sin \theta}{\cos^2 \theta} = \sqrt{2} \]

### Step 2: Multiply both sides by \( \cos^2 \theta \)
To eliminate the fraction, we multiply both sides by \( \cos^2 \theta \):
\[ \sin \theta = \sqrt{2} \cos^2 \theta \]

### Step 3: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Substituting this into our equation gives:
\[ \sin \theta = \sqrt{2} (1 - \sin^2 \theta) \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ \sqrt{2} \sin^2 \theta + \sin \theta - \sqrt{2} = 0 \]
This is a quadratic equation in terms of \( \sin \theta \).

### Step 5: Apply the quadratic formula
The quadratic formula is given by:
\[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = \sqrt{2} \), \( b = 1 \), and \( c = -\sqrt{2} \). Plugging in these values:
\[ \sin \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot \sqrt{2} \cdot (-\sqrt{2})}}{2 \cdot \sqrt{2}} \]
Calculating the discriminant:
\[ 1 + 8 = 9 \]
So we have:
\[ \sin \theta = \frac{-1 \pm 3}{2\sqrt{2}} \]

### Step 6: Solve for \( \sin \theta \)
This gives us two potential solutions:
1. \( \sin \theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \)
2. \( \sin \theta = \frac{-4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} \) (not valid since sine cannot be less than -1)

Thus, we have:
\[ \sin \theta = \frac{1}{\sqrt{2}} \]

### Step 7: Find the general solution for \( \theta \)
The angle \( \theta \) that satisfies \( \sin \theta = \frac{1}{\sqrt{2}} \) is:
\[ \theta = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
Thus, the solution for \( \theta \) is:
\[ \theta = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

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