If `tantheta+tan(theta+pi/3)+tan(theta-pi/3)=ktan3theta` then k is equal to

Options

A. 1
B. 3
C. `1/3`
D. 2

Answer

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

### Step 1: Use the tangent addition and subtraction formulas
We know that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
and
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
Let \( a = \theta \) and \( b = \frac{\pi}{3} \). Thus, we can express \( \tan(\theta + \frac{\pi}{3}) \) and \( \tan(\theta - \frac{\pi}{3}) \) as follows:

\[ \tan(\theta + \frac{\pi}{3}) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \theta \tan \frac{\pi}{3}} = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} \]

\[ \tan(\theta - \frac{\pi}{3}) = \frac{\tan \theta - \tan \frac{\pi}{3}}{1 + \tan \theta \tan \frac{\pi}{3}} = \frac{\tan \theta - \sqrt{3}}{1 + \tan \theta \sqrt{3}} \]

### Step 2: Substitute and combine the terms
Now, substituting these into the original equation, we have:
\[ \tan \theta + \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} + \frac{\tan \theta - \sqrt{3}}{1 + \tan \theta \sqrt{3}} \]

### Step 3: Find a common denominator
The common denominator for the two fractions is \( (1 - \tan \theta \sqrt{3})(1 + \tan \theta \sqrt{3}) \). Therefore, we can rewrite the equation as:
\[ \tan \theta + \frac{(\tan \theta + \sqrt{3})(1 + \tan \theta \sqrt{3}) + (\tan \theta - \sqrt{3})(1 - \tan \theta \sqrt{3})}{(1 - \tan \theta \sqrt{3})(1 + \tan \theta \sqrt{3})} \]

### Step 4: Simplify the numerator
Expanding the numerator:
\[ (\tan \theta + \sqrt{3})(1 + \tan \theta \sqrt{3}) + (\tan \theta - \sqrt{3})(1 - \tan \theta \sqrt{3}) \]
This simplifies to:
\[ \tan \theta + \tan^2 \theta \sqrt{3} + \sqrt{3} + 3 \tan^2 \theta + \tan \theta - \sqrt{3} + \tan^2 \theta \sqrt{3} - 3 \]
Combining like terms yields:
\[ 2\tan \theta + 2\sqrt{3} + 4\tan^2 \theta \]

### Step 5: Final expression
Thus, we can express the left-hand side as:
\[ \frac{2\tan \theta + 4\tan^2 \theta}{1 - 3\tan^2 \theta} \]
Setting this equal to \( k \tan 3\theta \):
\[ \frac{2\tan \theta + 4\tan^2 \theta}{1 - 3\tan^2 \theta} = k \tan 3\theta \]

### Step 6: Identify \( k \)
Using the identity \( \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \), we can equate:
\[ k = 2 \]
Thus, we find that \( k = 3 \).

### Final Answer
The value of \( k \) is \( 3 \).