Find the general value of `theta` from the equation `tantheta+tan2theta+tan3theta=tantheta.tan2theta.tan3theta.`

Options

A. `(npi)/(3)`
B. `(npi)/(4)`
C. `(npi)/(6)`
D. `(npi)/(2)`

Answer

To solve the equation \( \tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \cdot \tan 2\theta \cdot \tan 3\theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \cdot \tan 2\theta \cdot \tan 3\theta \]

### Step 2: Move \( \tan 3\theta \) to the right side
Rearranging gives us:
\[ \tan \theta + \tan 2\theta = \tan \theta \cdot \tan 2\theta \cdot \tan 3\theta - \tan 3\theta \]

### Step 3: Factor out \( \tan 3\theta \)
We can factor out \( \tan 3\theta \) from the right side:
\[ \tan \theta + \tan 2\theta = \tan 3\theta \left( \tan \theta \cdot \tan 2\theta - 1 \right) \]

### Step 4: Rearrange the equation
Next, we can rearrange the equation:
\[ \tan \theta + \tan 2\theta = -\tan 3\theta \cdot \left( 1 - \tan \theta \cdot \tan 2\theta \right) \]

### Step 5: Use the tangent addition formula
Using the tangent addition formula, we know that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Thus, we can express the left side as:
\[ \tan(2\theta + \theta) = \tan 3\theta \]
So we have:
\[ \tan 3\theta = -\tan 3\theta \]

### Step 6: Set up the equation
This implies:
\[ \tan 3\theta = \tan(-3\theta) \]

### Step 7: Solve for \( \theta \)
From the property of the tangent function, we know:
\[ 3\theta = n\pi - 3\theta \quad (n \in \mathbb{Z}) \]
Adding \( 3\theta \) to both sides gives:
\[ 6\theta = n\pi \]
Thus, we can solve for \( \theta \):
\[ \theta = \frac{n\pi}{6} \]

### Final Answer
The general value of \( \theta \) is:
\[ \theta = \frac{n\pi}{6}, \quad n \in \mathbb{Z} \]