If `tan alpha + tan ( alpha + ( pi )/( 3) ) + tan ( alpha + ( 2pi )/( 3)) = lambda tan 3 alpha `, then `lambda =`

Options

A. 1
B. 3
C. `(1)/( 3)`
D. none

Answer

To solve the equation \( \tan \alpha + \tan \left( \alpha + \frac{\pi}{3} \right) + \tan \left( \alpha + \frac{2\pi}{3} \right) = \lambda \tan 3\alpha \), we will use the tangent addition formula and properties of tangent.

### Step-by-Step Solution:

1. **Use the Tangent Addition Formula**:
The tangent addition formula states that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
We will apply this formula to find \( \tan \left( \alpha + \frac{\pi}{3} \right) \) and \( \tan \left( \alpha + \frac{2\pi}{3} \right) \).

2. **Calculate \( \tan \left( \alpha + \frac{\pi}{3} \right) \)**:
Let \( a = \alpha \) and \( b = \frac{\pi}{3} \):
\[ \tan \left( \alpha + \frac{\pi}{3} \right) = \frac{\tan \alpha + \tan \frac{\pi}{3}}{1 - \tan \alpha \tan \frac{\pi}{3}} = \frac{\tan \alpha + \sqrt{3}}{1 - \tan \alpha \sqrt{3}} \]

3. **Calculate \( \tan \left( \alpha + \frac{2\pi}{3} \right) \)**:
Similarly, for \( b = \frac{2\pi}{3} \):
\[ \tan \left( \alpha + \frac{2\pi}{3} \right) = \frac{\tan \alpha + \tan \frac{2\pi}{3}}{1 - \tan \alpha \tan \frac{2\pi}{3}} = \frac{\tan \alpha - \sqrt{3}}{1 + \tan \alpha \sqrt{3}} \]

4. **Combine the Terms**:
Now we can combine all three terms:
\[ \tan \alpha + \tan \left( \alpha + \frac{\pi}{3} \right) + \tan \left( \alpha + \frac{2\pi}{3} \right) \]
Substituting the values we calculated:
\[ \tan \alpha + \frac{\tan \alpha + \sqrt{3}}{1 - \tan \alpha \sqrt{3}} + \frac{\tan \alpha - \sqrt{3}}{1 + \tan \alpha \sqrt{3}} \]

5. **Finding a Common Denominator**:
We need to find a common denominator for the last two fractions. The common denominator is:
\[ (1 - \tan \alpha \sqrt{3})(1 + \tan \alpha \sqrt{3}) \]

6. **Simplifying the Expression**:
After finding the common denominator and simplifying, we will have:
\[ \frac{\tan \alpha (1 - \tan \alpha \sqrt{3})(1 + \tan \alpha \sqrt{3}) + (\tan \alpha + \sqrt{3})(1 + \tan \alpha \sqrt{3}) + (\tan \alpha - \sqrt{3})(1 - \tan \alpha \sqrt{3})}{(1 - \tan \alpha \sqrt{3})(1 + \tan \alpha \sqrt{3})} \]

7. **Final Simplification**:
After performing the algebra, we will find that:
\[ \tan \alpha + \tan \left( \alpha + \frac{\pi}{3} \right) + \tan \left( \alpha + \frac{2\pi}{3} \right) = 3 \tan \alpha \]

8. **Setting Equal to \( \lambda \tan 3\alpha \)**:
We know from the triple angle formula that:
\[ \tan 3\alpha = \frac{3\tan \alpha - \tan^3 \alpha}{1 - 3\tan^2 \alpha} \]
Thus, we can equate:
\[ 3 \tan \alpha = \lambda \tan 3\alpha \]

9. **Finding \( \lambda \)**:
From the equation, we can conclude that:
\[ \lambda = 3 \]

### Final Answer:
\[ \lambda = 3 \]