In `(0,6pi)`, the number of solutions of the equation `tantheta+tan2theta+tan3theta=tantheta*tan2theta*tan3theta` is/are

Options

A. `15`
B. `17`
C. `20`
D. `12`

Answer

To solve the equation \( \tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta \) in the interval \( (0, 6\pi) \), we can follow these steps:

### Step 1: Rearrange the Equation
We start with the equation:
\[ \tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta \]
Rearranging gives us:
\[ \tan \theta + \tan 2\theta = \tan \theta \tan 2\theta \tan 3\theta - \tan 3\theta \]
This can be rewritten as:
\[ \tan \theta + \tan 2\theta = \tan 3\theta (\tan \theta \tan 2\theta - 1) \]

### Step 2: Factor the Left Side
Now, we can express the left-hand side in terms of \( \tan 3\theta \):
\[ \frac{\tan \theta + \tan 2\theta}{\tan \theta \tan 2\theta - 1} = \tan 3\theta \]

### Step 3: Use the Identity for \( \tan 3\theta \)
Recall the identity for \( \tan 3\theta \):
\[ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]
Setting this equal to our expression gives:
\[ \frac{\tan \theta + \tan 2\theta}{\tan \theta \tan 2\theta - 1} = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]

### Step 4: Solve for \( \tan 3\theta = 0 \)
From the rearrangement, we find:
\[ \tan 3\theta = 0 \]
The solutions to \( \tan 3\theta = 0 \) occur at:
\[ 3\theta = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ \theta = \frac{n\pi}{3} \]

### Step 5: Find Values of \( n \) in the Interval \( (0, 6\pi) \)
To find the number of solutions in \( (0, 6\pi) \):
- The smallest \( n \) such that \( \theta > 0 \) is \( n = 1 \).
- The largest \( n \) such that \( \theta < 6\pi \) is \( n = 18 \) (since \( 3\theta < 18\pi \)).

Thus, \( n \) can take values from \( 1 \) to \( 18 \), giving us \( 18 - 1 + 1 = 18 \) solutions.

### Step 6: Exclude Endpoints
Since the interval is open at both ends, we do not include \( \theta = 0 \) or \( \theta = 6\pi \).

### Final Count of Solutions
Thus, the total number of solutions in the interval \( (0, 6\pi) \) is:
\[ \text{Number of solutions} = 18 \]