In an acute-angled triangle ABC, `tanA+tanB+tanC`

Options

A. `ge3`
B. `gesqrt3`
C. `ge3sqrt3`
D. none of these

Explanation

To find the value of \( \tan A + \tan B + \tan C \) in an acute-angled triangle \( ABC \), we can use the relationship between the tangent of the angles and the properties of triangles.

### Step-by-Step Solution:

1. **Understanding the Angles in Triangle**:
In any triangle, the sum of the angles is \( 180^\circ \). Therefore, we have:
\[ A + B + C = 180^\circ \]

2. **Using the Tangent Addition Formula**:
We know that:
\[ \tan(A + B + C) = \tan 180^\circ = 0 \]
Using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Thus, we can express \( \tan C \) as:
\[ \tan C = \tan(180^\circ - (A + B)) = -\tan(A + B) \]

3. **Substituting for \( \tan C \)**:
Since \( \tan C = -\tan(A + B) \), we can write:
\[ \tan C = -\frac{\tan A + \tan B}{1 - \tan A \tan B} \]

4. **Finding \( \tan A + \tan B + \tan C \)**:
Now substituting \( \tan C \) into the expression:
\[ \tan A + \tan B + \tan C = \tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \( x = \tan A + \tan B \). Then we have:
\[ x - \frac{x}{1 - \tan A \tan B} = x \left( 1 - \frac{1}{1 - \tan A \tan B} \right) \]
Simplifying gives:
\[ = x \left( \frac{-\tan A \tan B}{1 - \tan A \tan B} \right) \]

5. **Using AM-GM Inequality**:
By applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{\tan A + \tan B + \tan C}{3} \geq \sqrt[3]{\tan A \tan B \tan C} \]
This implies:
\[ \tan A + \tan B + \tan C \geq 3 \sqrt[3]{\tan A \tan B \tan C} \]

6. **Final Result**:
For an acute triangle, it can be shown that:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
Therefore, the value of \( \tan A + \tan B + \tan C \) simplifies to:
\[ \tan A + \tan B + \tan C = 3\sqrt{3} \]

### Conclusion:
Thus, the value of \( \tan A + \tan B + \tan C \) in an acute-angled triangle \( ABC \) is \( 3\sqrt{3} \).