If `tantheta+tan2theta+tan3theta=tanthetatan2thetatan3theta`, then the general value of `theta` is

Options

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

Answer

To solve the equation \( \tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta \), we can use a trigonometric identity related to the sum of tangents.

### Step-by-Step Solution:

1. **Recognize the Identity**:
We will use the identity for the tangent of a sum:
\[ \tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)} \]
In this case, let \( A = \theta \), \( B = 2\theta \), and \( C = 3\theta \).

2. **Apply the Identity**:
We can rewrite the left-hand side:
\[ \tan \theta + \tan 2\theta + \tan 3\theta = \tan(6\theta) \]
Thus, we have:
\[ \tan(6\theta) = \tan \theta \tan 2\theta \tan 3\theta \]

3. **Set Up the Equation**:
The equation now becomes:
\[ \tan(6\theta) = \tan \theta \tan 2\theta \tan 3\theta \]
Since we want to find when this equality holds, we will set the left-hand side to zero:
\[ \tan(6\theta) = 0 \]

4. **Solve for \( \theta \)**:
The tangent function is zero at integer multiples of \( \pi \):
\[ 6\theta = n\pi \quad \text{for } n \in \mathbb{Z} \]
Dividing both sides by 6 gives:
\[ \theta = \frac{n\pi}{6} \]

5. **General Solution**:
The general solution for \( \theta \) is:
\[ \theta = \frac{n\pi}{6}, \quad n \in \mathbb{Z} \]

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