If `A, B, C` are angles of `Delta ABC` and `tan A tan C = 3, tan B tan C = 6`, then

Options

A. `A=(pi)/(4)`
B. `tanA tanB=2`
C. `(tanA)/(tanC)=3`
D. `tanB=2tanA`

Answer

To solve the problem, we are given the following equations based on the angles of triangle ABC:

1. \( \tan A \tan C = 3 \) (Equation 1)
2. \( \tan B \tan C = 6 \) (Equation 2)

Since \( A + B + C = 180^\circ \), we can express \( B \) in terms of \( A \) and \( C \):
\[ B = 180^\circ - A - C \]

Using the tangent subtraction formula:
\[ \tan(180^\circ - x) = -\tan x \]
Thus,
\[ \tan B = -\tan(A + C) = -\frac{\tan A + \tan C}{1 - \tan A \tan C} \]

Now substituting \( \tan A \tan C = 3 \) into the equation for \( \tan B \):
\[ \tan B = -\frac{\tan A + \tan C}{1 - 3} \]
\[ \tan B = -\frac{\tan A + \tan C}{-2} = \frac{\tan A + \tan C}{2} \]

Now we can substitute \( \tan B \) into Equation 2:
\[ \frac{\tan A + \tan C}{2} \tan C = 6 \]
Multiplying both sides by 2:
\[ (\tan A + \tan C) \tan C = 12 \]

Now we can express \( \tan A \) in terms of \( \tan C \):
From Equation 1, we have:
\[ \tan A = \frac{3}{\tan C} \]

Substituting this into the equation:
\[ \left(\frac{3}{\tan C} + \tan C\right) \tan C = 12 \]
\[ 3 + \tan^2 C = 12 \]
\[ \tan^2 C = 12 - 3 = 9 \]
\[ \tan C = 3 \quad \text{(since tangent is positive in a triangle)} \]

Now substituting \( \tan C = 3 \) back into Equation 1 to find \( \tan A \):
\[ \tan A \cdot 3 = 3 \implies \tan A = 1 \]
Thus,
\[ A = 45^\circ \]

Now substituting \( \tan C = 3 \) into Equation 2 to find \( \tan B \):
\[ \tan B \cdot 3 = 6 \implies \tan B = 2 \]
Thus,
\[ B = \tan^{-1}(2) \]

Finally, we can find \( C \):
\[ C = 180^\circ - A - B = 180^\circ - 45^\circ - \tan^{-1}(2) \]

### Summary of Results:
- \( A = 45^\circ \)
- \( B = \tan^{-1}(2) \)
- \( C = 180^\circ - 45^\circ - \tan^{-1}(2) \)