`tan""(2pi)/(5)-tan""(pi)/(15)-sqrt3tan""(2pi)/(5)tan""(pi)/(15)` is equal to

Options

A. `-sqrt3`
B. `1//sqrt3`
C. 1
D. `sqrt3`

Answer

To solve the expression \( \tan\left(\frac{2\pi}{5}\right) - \tan\left(\frac{\pi}{15}\right) - \sqrt{3} \tan\left(\frac{2\pi}{5}\right) \tan\left(\frac{\pi}{15}\right) \), we will follow these steps:

### Step 1: Define the angles
Let:
- \( a = \tan\left(\frac{2\pi}{5}\right) \)
- \( b = \tan\left(\frac{\pi}{15}\right) \)

Then, we can rewrite the expression as:
\[ a - b - \sqrt{3}ab \]

### Step 2: Use the tangent subtraction formula
We know that:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

In our case, we can set \( A = \frac{2\pi}{5} \) and \( B = \frac{\pi}{15} \). Thus:
\[ \tan\left(\frac{2\pi}{5} - \frac{\pi}{15}\right) = \frac{a - b}{1 + ab} \]

### Step 3: Calculate \( A - B \)
To find \( A - B \):
\[ \frac{2\pi}{5} - \frac{\pi}{15} = \frac{6\pi}{15} - \frac{\pi}{15} = \frac{5\pi}{15} = \frac{\pi}{3} \]

### Step 4: Find \( \tan\left(\frac{\pi}{3}\right) \)
We know that:
\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

### Step 5: Substitute back into the tangent formula
From the tangent subtraction formula:
\[ \sqrt{3} = \frac{a - b}{1 + ab} \]

### Step 6: Rearranging the equation
Now, we can rearrange this equation:
\[ \sqrt{3}(1 + ab) = a - b \]
This leads to:
\[ a - b - \sqrt{3}ab = 0 \]

### Step 7: Conclusion
Thus, we find that:
\[ \tan\left(\frac{2\pi}{5}\right) - \tan\left(\frac{\pi}{15}\right) - \sqrt{3} \tan\left(\frac{2\pi}{5}\right) \tan\left(\frac{\pi}{15}\right) = 0 \]

### Final Answer
The expression evaluates to \( 0 \).

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