In a `Delta ABC`, if `cosA cos B cos C= (sqrt3-1)/(8) and sin A sin B sin C= (3+ sqrt3)/(8)`, then
The value of `tan A + tan B + tan C` is

Options

A. `(3+ sqrt3)/(sqrt3-1)`
B. `(sqrt3 +4)/(sqrt3-1)`
C. `(6-sqrt3)/(sqrt3-1)`
D. `(sqrt3+ sqrt2)/(sqrt3-1)`

Explanation

To solve the problem, we need to find the value of \( \tan A + \tan B + \tan C \) given the equations:

1. \( \cos A \cos B \cos C = \frac{\sqrt{3} - 1}{8} \)
2. \( \sin A \sin B \sin C = \frac{3 + \sqrt{3}}{8} \)

### Step-by-Step Solution:

**Step 1: Write down the given equations.**

We have:
\[ \cos A \cos B \cos C = \frac{\sqrt{3} - 1}{8} \quad \text{(Equation 1)} \]
\[ \sin A \sin B \sin C = \frac{3 + \sqrt{3}}{8} \quad \text{(Equation 2)} \]

**Step 2: Divide Equation 2 by Equation 1.**

We divide the second equation by the first:
\[ \frac{\sin A \sin B \sin C}{\cos A \cos B \cos C} = \frac{\frac{3 + \sqrt{3}}{8}}{\frac{\sqrt{3} - 1}{8}} \]
This simplifies to:
\[ \frac{\sin A \sin B \sin C}{\cos A \cos B \cos C} = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \]

**Step 3: Use the identity \( \frac{\sin \theta}{\cos \theta} = \tan \theta \).**

Thus, we can write:
\[ \tan A \tan B \tan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \quad \text{(Equation 3)} \]

**Step 4: Use the identity for angles in a triangle.**

In triangle \( ABC \), we know:
\[ A + B + C = 180^\circ \quad \Rightarrow \quad C = 180^\circ - (A + B) \]

**Step 5: Express \( \tan C \) in terms of \( A \) and \( B \).**

Using the identity:
\[ \tan(180^\circ - \theta) = -\tan \theta \]
We can write:
\[ \tan C = -\tan(A + B) \]

**Step 6: Use the formula for \( \tan(A + B) \).**

Using the formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Thus:
\[ \tan C = -\frac{\tan A + \tan B}{1 - \tan A \tan B} \]

**Step 7: Substitute \( \tan C \) into the equation.**

Now we can express \( \tan A + \tan B + \tan C \):
\[ \tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

**Step 8: Factor out \( \tan A + \tan B \).**

Let \( x = \tan A + \tan B \). Then:
\[ x - \frac{x}{1 - \tan A \tan B} = x \left(1 - \frac{1}{1 - \tan A \tan B}\right) \]
This simplifies to:
\[ x \left(\frac{-\tan A \tan B}{1 - \tan A \tan B}\right) \]

**Step 9: Set up the equation.**

From Equation 3, we have:
\[ \tan A \tan B \tan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \]
Substituting \( \tan C \):
\[ \tan A \tan B \left(-\frac{\tan A + \tan B}{1 - \tan A \tan B}\right) = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \]

**Step 10: Solve for \( \tan A + \tan B + \tan C \).**

Finally, we can conclude that:
\[ \tan A + \tan B + \tan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \]

### Final Answer:
\[ \tan A + \tan B + \tan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \]