General solution of `tan theta+tan 4 theta+tan 7 theta=tan theta tan 4 theta tan 7 theta` is

Options

A. `theta=n pi//12`, where `n in Z`
B. `theta=n pi//9`, where `n in Z`
C. `theta = npi+pi//12`, where `n in Z`
D. none of these

Explanation

To solve the equation \( \tan \theta + \tan 4\theta + \tan 7\theta = \tan \theta \tan 4\theta \tan 7\theta \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \tan \theta + \tan 4\theta + \tan 7\theta = \tan \theta \tan 4\theta \tan 7\theta \]
We can rearrange it by moving all terms to one side:
\[ \tan \theta + \tan 4\theta + \tan 7\theta - \tan \theta \tan 4\theta \tan 7\theta = 0 \]

### Step 2: Grouping Terms
Next, we can group the terms involving \( \tan 7\theta \) to one side:
\[ \tan 7\theta = \tan \theta + \tan 4\theta - \tan \theta \tan 4\theta \tan 7\theta \]
This can be rewritten as:
\[ \tan 7\theta = \tan \theta + \tan 4\theta - \tan \theta \tan 4\theta \tan 7\theta \]

### Step 3: Using the Tangent Addition Formula
Recall the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
We can express \( \tan 7\theta \) in terms of \( \tan \theta \) and \( \tan 4\theta \):
\[ \tan 7\theta = \tan(\theta + 4\theta) = \frac{\tan \theta + \tan 4\theta}{1 - \tan \theta \tan 4\theta} \]

### Step 4: Setting Up the Equation
Now we can set the two expressions for \( \tan 7\theta \) equal to each other:
\[ \frac{\tan \theta + \tan 4\theta}{1 - \tan \theta \tan 4\theta} = \tan \theta + \tan 4\theta - \tan \theta \tan 4\theta \tan 7\theta \]

### Step 5: Simplifying the Equation
Cross-multiplying gives:
\[ (\tan \theta + \tan 4\theta)(1 - \tan \theta \tan 4\theta) = \tan \theta + \tan 4\theta - \tan \theta \tan 4\theta \tan 7\theta \]
This leads to:
\[ \tan \theta + \tan 4\theta - \tan \theta \tan 4\theta (\tan \theta + \tan 4\theta) = 0 \]

### Step 6: Factorization
Factoring out \( \tan \theta + \tan 4\theta \):
\[ (\tan \theta + \tan 4\theta)(1 - \tan \theta \tan 4\theta) = 0 \]
This gives us two cases:
1. \( \tan \theta + \tan 4\theta = 0 \)
2. \( 1 - \tan \theta \tan 4\theta = 0 \)

### Step 7: Solving Each Case
**Case 1:** \( \tan \theta + \tan 4\theta = 0 \)
\[ \tan 4\theta = -\tan \theta \]
This implies:
\[ 4\theta = \theta + (2n + 1)\frac{\pi}{2} \quad (n \in \mathbb{Z}) \]
Simplifying gives:
\[ 3\theta = (2n + 1)\frac{\pi}{2} \implies \theta = \frac{(2n + 1)\pi}{6} \]

**Case 2:** \( 1 - \tan \theta \tan 4\theta = 0 \)
\[ \tan \theta \tan 4\theta = 1 \]
This implies:
\[ 4\theta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Thus:
\[ \theta = \frac{\pi}{8} + \frac{n\pi}{4} \]

### Final General Solution
Combining both cases, we have:
1. \( \theta = \frac{(2n + 1)\pi}{6} \)
2. \( \theta = \frac{\pi}{8} + \frac{n\pi}{4} \)