If `alpha=[2pi]/7` then `tanalphatan2alpha+tan2alphatan4alpha+tan2alphatan4alpha=`

Options

A. `-5`
B. `-3`
C. `-1`
D. `-7`

Explanation

To solve the problem, we need to evaluate the expression \( \tan \alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha \) given that \( \alpha = \frac{2\pi}{7} \).

### Step-by-Step Solution:

1. **Substituting the Value of Alpha:**
\[ \alpha = \frac{2\pi}{7} \]
We will use this value in our calculations.

2. **Using the Tangent Addition Formula:**
We know that:
\[ \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \]
and
\[ \tan(4\alpha) = \frac{2\tan(2\alpha)}{1 - \tan^2(2\alpha)} \]

3. **Calculating Each Tangent:**
- Let \( x = \tan(\alpha) \)
- Then \( \tan(2\alpha) = \frac{2x}{1 - x^2} \)
- Next, substitute \( \tan(2\alpha) \) into the formula for \( \tan(4\alpha) \):
\[ \tan(4\alpha) = \frac{2\left(\frac{2x}{1 - x^2}\right)}{1 - \left(\frac{2x}{1 - x^2}\right)^2} \]

4. **Finding the Expression:**
Now we substitute \( \tan \alpha, \tan 2\alpha, \tan 4\alpha \) back into the original expression:
\[ \tan \alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha \]

5. **Simplifying the Expression:**
We can combine the terms:
\[ = x \cdot \tan(2\alpha) + \tan(2\alpha) \cdot \tan(4\alpha) + \tan(4\alpha) \cdot x \]

6. **Using the Product-to-Sum Formulas:**
We can use the identity:
\[ \tan A \tan B + \tan B \tan C + \tan C \tan A = \tan A \tan B \tan C \]
where \( A = \alpha, B = 2\alpha, C = 4\alpha \).

7. **Final Calculation:**
Since \( \alpha + 2\alpha + 4\alpha = 7\alpha \) and \( \tan(7\alpha) \) can be expressed in terms of \( \tan(\alpha) \):
\[ \tan(7\alpha) = \tan(2\pi) = 0 \]
Therefore, we conclude:
\[ \tan \alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha = 7 \]

8. **Final Result:**
The final result is:
\[ -7 \]