Which one of the following in one of the solutions of the equation tan 2`theta`, tan`theta` = 1?

Options

A. `(pi)/(12)`
B. `(pi)/(6)`
C. `(pi)/(4)`
D. `(pi)/(3)`

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
The double angle formula for tangent is given by:
\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
Substituting this into the equation, we have:
\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot \tan \theta = 1 \]

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

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

### Step 4: Rearrange the equation
Rearranging the equation, we get:
\[ 2 \tan^2 \theta + \tan^2 \theta = 1 \]
\[ 3 \tan^2 \theta = 1 \]

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

### Step 6: Solve for \( \tan \theta \)
Taking the square root of both sides:
\[ \tan \theta = \pm \frac{1}{\sqrt{3}} \]

### Step 7: Find the angles corresponding to \( \tan \theta = \frac{1}{\sqrt{3}} \)
The angles where \( \tan \theta = \frac{1}{\sqrt{3}} \) are:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 8: Find the angles corresponding to \( \tan \theta = -\frac{1}{\sqrt{3}} \)
The angles where \( \tan \theta = -\frac{1}{\sqrt{3}} \) are:
\[ \theta = \frac{5\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 9: Identify the solutions from the options
Now, we check the options:
1. \( \frac{\pi}{12} \)
2. \( \frac{\pi}{6} \) (This is a solution)
3. \( \frac{\pi}{4} \)
4. \( \frac{\pi}{3} \)

Thus, one of the solutions to the equation \( \tan 2\theta \cdot \tan \theta = 1 \) is:
\[ \theta = \frac{\pi}{6} \]