If: `tantheta + tan 2theta + tan 3theta = tan theta.tan2theta.tan 3theta`, then: `theta`=

Options

A. `npi`
B. `n pi/6`
C. `npi +- pi/3`
D. `n pi/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 will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation to isolate the terms involving \( \tan 3\theta \):
\[ \tan \theta + \tan 2\theta = \tan \theta \cdot \tan 2\theta \cdot \tan 3\theta - \tan 3\theta \]
This simplifies to:
\[ \tan \theta + \tan 2\theta - \tan 3\theta = \tan \theta \cdot \tan 2\theta \cdot \tan 3\theta \]

### Step 2: Factoring Out \( \tan 3\theta \)
Now, we can factor out \( \tan 3\theta \) from the right-hand side:
\[ \tan \theta + \tan 2\theta = \tan 3\theta (\tan \theta \cdot \tan 2\theta - 1) \]

### Step 3: Using the Tangent Addition Formula
Recall the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Thus, we can rewrite the left-hand side:
\[ \tan(\theta + 2\theta) = \tan(3\theta) \]
This gives us:
\[ \tan(\theta + 2\theta) = \tan 3\theta \]

### Step 4: Setting the Angles Equal
Since the tangents are equal, we can set the angles equal to each other (considering the periodic nature of the tangent function):
\[ \theta + 2\theta = 3\theta = n\pi \quad \text{(where \( n \) is any integer)} \]

### Step 5: Solving for \( \theta \)
From the equation \( 3\theta = n\pi \), we can solve for \( \theta \):
\[ \theta = \frac{n\pi}{3} \]

### Final Answer
Thus, the solution for \( \theta \) is:
\[ \theta = \frac{n\pi}{3}, \quad n \in \mathbb{Z} \]

---