If `tan 3 theta + tan theta =2 tan 2 theta`, then `theta` is equal to `(n in Z)`

Options

A. `n pi`
B. `(n pi)/4`
C. `(n pi)/2`
D. none of these

Explanation

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

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
Substituting this into the equation gives:
\[ \tan 3\theta + \tan \theta = 2 \left(\frac{2 \tan \theta}{1 - \tan^2 \theta}\right) \]

### Step 2: Express \(\tan 3\theta\) in terms of \(\tan \theta\)
Using the identity for \(\tan 3\theta\):
\[ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]
We can substitute this into our equation:
\[ \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} + \tan \theta = \frac{4\tan \theta}{1 - \tan^2 \theta} \]

### Step 3: Combine the left-hand side
Combine the terms on the left-hand side:
\[ \frac{3\tan \theta - \tan^3 \theta + \tan \theta(1 - 3\tan^2 \theta)}{1 - 3\tan^2 \theta} = \frac{4\tan \theta}{1 - \tan^2 \theta} \]
This simplifies to:
\[ \frac{4\tan \theta - \tan^3 \theta - 3\tan^3 \theta}{1 - 3\tan^2 \theta} = \frac{4\tan \theta}{1 - \tan^2 \theta} \]
Thus:
\[ \frac{4\tan \theta - 4\tan^3 \theta}{1 - 3\tan^2 \theta} = \frac{4\tan \theta}{1 - \tan^2 \theta} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ (4\tan \theta - 4\tan^3 \theta)(1 - \tan^2 \theta) = 4\tan \theta(1 - 3\tan^2 \theta) \]

### Step 5: Expand and simplify
Expanding both sides:
\[ 4\tan \theta - 4\tan^3 \theta - 4\tan^3 \theta + 4\tan^5 \theta = 4\tan \theta - 12\tan^3 \theta \]
This simplifies to:
\[ 4\tan^5 \theta - 4\tan^3 \theta = -12\tan^3 \theta \]
Rearranging gives:
\[ 4\tan^5 \theta + 8\tan^3 \theta = 0 \]

### Step 6: Factor out common terms
Factoring out \(4\tan^3 \theta\):
\[ 4\tan^3 \theta(\tan^2 \theta + 2) = 0 \]

### Step 7: Solve for \(\tan \theta\)
Setting each factor to zero:
1. \(4\tan^3 \theta = 0\) implies \(\tan \theta = 0\)
2. \(\tan^2 \theta + 2 = 0\) has no real solutions.

### Step 8: Find values for \(\theta\)
From \(\tan \theta = 0\), we have:
\[ \theta = n\pi, \quad n \in \mathbb{Z} \]

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