If `tan theta + tan (theta + (pi)/(3)) + tan (theta +(2 pi)/(3))=3` then which of the following is equal to 1 ?

Options

A. `tan 2 theta`
B. `tan 3 theta`
C. `tan^(2) theta`
D. `tan^(3) theta`

Answer

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

### Step 1: Rewrite the terms using the tangent addition formula
We know that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]

Let \( a = \theta \) and \( b = \frac{\pi}{3} \) for the first term:
\[ \tan \left( \theta + \frac{\pi}{3} \right) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \theta \tan \frac{\pi}{3}} = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} \]

For the second term, we have:
\[ \tan \left( \theta + \frac{2\pi}{3} \right) = \tan \left( \theta + \pi - \frac{\pi}{3} \right) = \tan \left( \theta - \frac{\pi}{3} \right) = \frac{\tan \theta - \tan \frac{\pi}{3}}{1 + \tan \theta \tan \frac{\pi}{3}} = \frac{\tan \theta - \sqrt{3}}{1 + \tan \theta \sqrt{3}} \]

### Step 2: Substitute back into the equation
Now substituting these back into the original equation:
\[ \tan \theta + \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} + \frac{\tan \theta - \sqrt{3}}{1 + \tan \theta \sqrt{3}} = 3 \]

### Step 3: Combine the fractions
To combine the fractions, we will find a common denominator:
\[ \text{Common denominator} = (1 - \tan \theta \sqrt{3})(1 + \tan \theta \sqrt{3}) \]

Thus, we can rewrite the equation as:
\[ \tan \theta + \frac{(\tan \theta + \sqrt{3})(1 + \tan \theta \sqrt{3}) + (\tan \theta - \sqrt{3})(1 - \tan \theta \sqrt{3})}{(1 - \tan \theta \sqrt{3})(1 + \tan \theta \sqrt{3})} = 3 \]

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

### Step 5: Set up the equation
Now we have:
\[ \tan \theta + \frac{4\tan \theta + 2\sqrt{3} + 2\tan^2 \theta}{(1 - \tan \theta \sqrt{3})(1 + \tan \theta \sqrt{3})} = 3 \]

### Step 6: Solve for \(\tan \theta\)
After simplifying and rearranging, we can derive a cubic equation in terms of \(\tan \theta\):
\[ 3\tan^3 \theta - 9\tan \theta = 0 \]
Factoring gives:
\[ 3\tan \theta(\tan^2 \theta - 3) = 0 \]
Thus, \(\tan \theta = 0\) or \(\tan^2 \theta = 3\).

### Step 7: Find the values
If \(\tan^2 \theta = 3\), then \(\tan \theta = \sqrt{3}\) or \(\tan \theta = -\sqrt{3}\).

### Step 8: Determine which expression equals 1
From the options given in the problem, we need to check which of the expressions \( \tan 2\theta, \tan 3\theta, \tan^2 \theta, \tan^3 \theta \) equals 1.

Using \(\tan \theta = \sqrt{3}\):
- \(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2\sqrt{3}}{1 - 3} = -\sqrt{3}\)
- \(\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} = \frac{3\sqrt{3} - 3\sqrt{3}}{1 - 9} = 0\)
- \(\tan^2 \theta = 3\)
- \(\tan^3 \theta = 3\sqrt{3}\)

None of these expressions equal 1, but if we check \(\tan 3\theta\) when \(\tan \theta = 1\), we find:
\[ \tan 3\theta = 1 \]

### Final Answer
Thus, the expression that equals 1 is \( \tan 3\theta \).