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 tanC ` is

Options

A. `5-4sqrt3`
B. `5+4sqrt3`
C. `3+2sqrt3`
D. `3-2sqrt3`

Explanation

To find the value of \( \tan A \tan B \tan C \) given that \( \cos A \cos B \cos C = \frac{\sqrt{3}-1}{8} \) and \( \sin A \sin B \sin C = \frac{3+\sqrt{3}}{8} \), we can follow these steps:

### Step 1: Write the expression for \( \tan A \tan B \tan C \)

We know that:
\[ \tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B}, \quad \tan C = \frac{\sin C}{\cos C} \]
Thus,
\[ \tan A \tan B \tan C = \frac{\sin A \sin B \sin C}{\cos A \cos B \cos C} \]

### Step 2: Substitute the given values

Substituting the values we have:
\[ \tan A \tan B \tan C = \frac{\sin A \sin B \sin C}{\cos A \cos B \cos C} = \frac{\frac{3+\sqrt{3}}{8}}{\frac{\sqrt{3}-1}{8}} \]

### Step 3: Simplify the expression

The \( \frac{8}{8} \) cancels out:
\[ \tan A \tan B \tan C = \frac{3+\sqrt{3}}{\sqrt{3}-1} \]

### Step 4: Rationalize the denominator

To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3}+1 \):
\[ \tan A \tan B \tan C = \frac{(3+\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \]

### Step 5: Calculate the denominator

Using the difference of squares:
\[ (\sqrt{3}-1)(\sqrt{3}+1) = 3 - 1 = 2 \]

### Step 6: Calculate the numerator

Now, calculate the numerator:
\[ (3+\sqrt{3})(\sqrt{3}+1) = 3\sqrt{3} + 3 + \sqrt{3}\cdot\sqrt{3} + \sqrt{3} = 3\sqrt{3} + 3 + 3 + \sqrt{3} = 4\sqrt{3} + 6 \]

### Step 7: Combine the results

Thus,
\[ \tan A \tan B \tan C = \frac{4\sqrt{3} + 6}{2} = 2\sqrt{3} + 3 \]

### Final Answer

The value of \( \tan A \tan B \tan C \) is:
\[ \boxed{2\sqrt{3} + 3} \]