If `tan theta + tan (pi/3 +theta) + tan (frac{2pi}{3}+ theta) = k tan 3theta` then k is equal to

Options

A. 1
B. `1/3`
C. 0
D. 3

Answer

To solve the equation \( \tan \theta + \tan \left( \frac{\pi}{3} + \theta \right) + \tan \left( \frac{2\pi}{3} + \theta \right) = k \tan 3\theta \), we will follow these steps:

### Step 1: Use the tangent addition formula
We know that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
We will apply this formula to find \( \tan \left( \frac{\pi}{3} + \theta \right) \) and \( \tan \left( \frac{2\pi}{3} + \theta \right) \).

### Step 2: Calculate \( \tan \left( \frac{\pi}{3} + \theta \right) \)
Using the formula:
\[ \tan \left( \frac{\pi}{3} + \theta \right) = \frac{\tan \frac{\pi}{3} + \tan \theta}{1 - \tan \frac{\pi}{3} \tan \theta} \]
Since \( \tan \frac{\pi}{3} = \sqrt{3} \), we have:
\[ \tan \left( \frac{\pi}{3} + \theta \right) = \frac{\sqrt{3} + \tan \theta}{1 - \sqrt{3} \tan \theta} \]

### Step 3: Calculate \( \tan \left( \frac{2\pi}{3} + \theta \right) \)
Using the formula:
\[ \tan \left( \frac{2\pi}{3} + \theta \right) = \frac{\tan \frac{2\pi}{3} + \tan \theta}{1 - \tan \frac{2\pi}{3} \tan \theta} \]
Since \( \tan \frac{2\pi}{3} = -\sqrt{3} \), we have:
\[ \tan \left( \frac{2\pi}{3} + \theta \right) = \frac{-\sqrt{3} + \tan \theta}{1 + \sqrt{3} \tan \theta} \]

### Step 4: Combine the three tangent terms
Now we can write:
\[ LHS = \tan \theta + \tan \left( \frac{\pi}{3} + \theta \right) + \tan \left( \frac{2\pi}{3} + \theta \right) \]
Substituting the values we calculated:
\[ LHS = \tan \theta + \frac{\sqrt{3} + \tan \theta}{1 - \sqrt{3} \tan \theta} + \frac{-\sqrt{3} + \tan \theta}{1 + \sqrt{3} \tan \theta} \]

### Step 5: Simplify the expression
To combine these fractions, we will find a common denominator:
\[ LHS = \tan \theta + \frac{(\sqrt{3} + \tan \theta)(1 + \sqrt{3} \tan \theta) + (-\sqrt{3} + \tan \theta)(1 - \sqrt{3} \tan \theta)}{(1 - \sqrt{3} \tan \theta)(1 + \sqrt{3} \tan \theta)} \]

### Step 6: Expand and simplify the numerator
Expanding the numerator:
\[ (\sqrt{3} + \tan \theta)(1 + \sqrt{3} \tan \theta) = \sqrt{3} + 3 \tan^2 \theta + \tan \theta \]
\[ (-\sqrt{3} + \tan \theta)(1 - \sqrt{3} \tan \theta) = -\sqrt{3} + 3 \tan^2 \theta - \tan \theta \]
Combining these gives:
\[ 2\sqrt{3} + 6\tan^2 \theta \]

### Step 7: Final expression and equate to \( k \tan 3\theta \)
Now we have:
\[ LHS = \tan \theta + \frac{2\sqrt{3} + 6\tan^2 \theta}{1 - 3\tan^2 \theta} \]
We know that:
\[ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]
Setting \( LHS = k \tan 3\theta \) and simplifying gives us \( k = 3 \).

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{3} \]