If `"tan" 2 theta "tan" theta =1, "then" theta =`

Options

A. `n pi + (pi)/(6), n in Z`
B. `n pi +- (pi)/(6), n in Z`
C. `2n pi +- (pi)/(6), n in Z`
D. none of these

Answer

To solve the equation \( \tan 2\theta \cdot \tan \theta = 1 \), we can follow these steps:

### Step 1: Use the double angle formula for tangent
We know that:
\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
Substituting this into the equation gives us:
\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot \tan \theta = 1 \]

### Step 2: Simplify the equation
This simplifies to:
\[ \frac{2 \tan^2 \theta}{1 - \tan^2 \theta} = 1 \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 2 \tan^2 \theta = 1 - \tan^2 \theta \]

### Step 4: Rearrange the equation
Rearranging this equation leads to:
\[ 2 \tan^2 \theta + \tan^2 \theta = 1 \]
\[ 3 \tan^2 \theta = 1 \]

### Step 5: Solve for \( \tan^2 \theta \)
Dividing both sides by 3 gives:
\[ \tan^2 \theta = \frac{1}{3} \]

### Step 6: Take the square root to find \( \tan \theta \)
Taking the square root of both sides, we find:
\[ \tan \theta = \pm \frac{1}{\sqrt{3}} \]

### Step 7: Identify the angles corresponding to \( \tan \theta \)
We know that:
\[ \tan \theta = \frac{1}{\sqrt{3}} \quad \text{corresponds to} \quad \theta = \frac{\pi}{6} + n\pi \]
\[ \tan \theta = -\frac{1}{\sqrt{3}} \quad \text{corresponds to} \quad \theta = -\frac{\pi}{6} + n\pi \]

### Step 8: Write the general solution
Thus, the general solutions for \( \theta \) can be expressed as:
\[ \theta = n\pi + \frac{\pi}{6} \quad \text{or} \quad \theta = n\pi - \frac{\pi}{6} \]

### Final Answer
The solutions for \( \theta \) can be summarized as:
\[ \theta = n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z} \]