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 \( \tan(2\theta) \tan(\theta) = 1 \), we will 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 gives:
\[ \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \tan(\theta) = 1 \]

### Step 2: Simplify the equation
This simplifies to:
\[ \frac{2\tan^2(\theta)}{1 - \tan^2(\theta)} = 1 \]
Now, cross-multiply to eliminate the fraction:
\[ 2\tan^2(\theta) = 1 - \tan^2(\theta) \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ 2\tan^2(\theta) + \tan^2(\theta) = 1 \]
\[ 3\tan^2(\theta) = 1 \]

### Step 4: Solve for \(\tan^2(\theta)\)
Dividing both sides by 3, we find:
\[ \tan^2(\theta) = \frac{1}{3} \]

### Step 5: Take the square root
Taking the square root of both sides gives:
\[ \tan(\theta) = \pm \frac{1}{\sqrt{3}} \]

### Step 6: Find the general solutions
The general solution for \(\tan(\theta) = \frac{1}{\sqrt{3}}\) is:
\[ \theta = n\pi + \frac{\pi}{6} \]
And for \(\tan(\theta) = -\frac{1}{\sqrt{3}}\), the general solution is:
\[ \theta = n\pi - \frac{\pi}{6} \]

### Final Solution
Combining both solutions, we have:
\[ \theta = n\pi \pm \frac{\pi}{6} \]
where \( n \) is any integer.

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