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 respective values of `tan A, tan B and tanC` are

Options

A. `1, sqrt3, sqrt2`
B. `1, sqrt3, 2`
C. `1, 2, sqrt3`
D. `1, sqrt3, 2+sqrt3`

Explanation

To solve the problem, we need to find the values of \( \tan A, \tan B, \) and \( \tan C \) 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 1: Find \( \tan A \tan B \tan C \)

We start by dividing 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:

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

### Step 2: Simplify the Right Side

To simplify \( \frac{3 + \sqrt{3}}{\sqrt{3} - 1} \), we can multiply the numerator and the denominator by the conjugate of the denominator:

\[ \frac{(3 + \sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(3\sqrt{3} + 3 + 3 + \sqrt{3})}{3 - 1} = \frac{4 + 4\sqrt{3}}{2} = 2 + 2\sqrt{3} \]

Thus, we have:

\[ \tan A \tan B \tan C = 2 + 2\sqrt{3} \]

### Step 3: Find \( \tan A + \tan B + \tan C \)

Using the identity for the sum of angles in a triangle, we know:

\[ \tan(A + B + C) = \tan(\pi) = 0 \]

Using the formula for the tangent of a sum:

\[ \tan A + \tan B + \tan C - \tan A \tan B \tan C = 0 \]

This gives us:

\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]

Thus:

\[ \tan A + \tan B + \tan C = 2 + 2\sqrt{3} \]

### Step 4: Find \( \tan A \tan B + \tan B \tan C + \tan C \tan A \)

Using the cosine identity:

\[ \cos(A + B + C) = -1 \]

We can express this in terms of tangents:

\[ \cos A \cos B \cos C (1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)) = -1 \]

Substituting the known value of \( \cos A \cos B \cos C \):

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

Rearranging gives:

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

Thus:

\[ \tan A \tan B + \tan B \tan C + \tan C \tan A = 1 + \frac{8}{\sqrt{3} - 1} \]

### Step 5: Form the Cubic Equation

Now we can form the cubic equation whose roots are \( \tan A, \tan B, \tan C \):

\[ x^3 - (2 + 2\sqrt{3})x^2 + (1 + \frac{8}{\sqrt{3} - 1})x - (2 + 2\sqrt{3}) = 0 \]

### Conclusion

The values of \( \tan A, \tan B, \tan C \) can be determined from the roots of this cubic equation.