If `tan2thetatantheta=1`, then the general value of `theta` is

Options

A. `(n+(1)/(2))(pi)/(3)`
B. `(n+(1)/(2))pi`
C. `(2npm(1)/(2))(pi)/(3)`
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
We know that:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
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:
\[ \frac{2\tan^2(\theta)}{1 - \tan^2(\theta)} = 1 \]

### 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 the equation results in:
\[ 2\tan^2(\theta) + \tan^2(\theta) = 1 \]
\[ 3\tan^2(\theta) = 1 \]

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

### Step 7: Determine the general solution for \(\theta\)
The angles for which \(\tan(\theta) = \frac{1}{\sqrt{3}}\) are:
\[ \theta = \frac{\pi}{6} + n\pi \quad \text{(for positive)} \]
And for \(\tan(\theta) = -\frac{1}{\sqrt{3}}\):
\[ \theta = \frac{5\pi}{6} + n\pi \quad \text{(for negative)} \]

### Final General Solution
Thus, the general solution for \(\theta\) is:
\[ \theta = n\pi + \frac{\pi}{6} \quad \text{or} \quad \theta = n\pi + \frac{5\pi}{6}, \quad n \in \mathbb{Z} \]