Find the value of `tanA+tanB+tanC`, if `A+B+C=pi`:

Options

A. `tanAtanB tanC`
B. 1
C. 0
D. `cotAcotB cotC`

Answer

To find the value of \( \tan A + \tan B + \tan C \) given that \( A + B + C = \pi \), we can use the trigonometric identity involving the tangent of angles.

### Step-by-Step Solution:

1. **Use the identity for tangent of a sum**:
Since \( A + B + C = \pi \), we can express \( C \) as \( C = \pi - (A + B) \).

2. **Apply the tangent function**:
We know that:
\[ \tan C = \tan(\pi - (A + B)) = -\tan(A + B) \]

3. **Use the tangent addition formula**:
The tangent of a sum can be expressed as:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Thus, we can rewrite \( \tan C \):
\[ \tan C = -\frac{\tan A + \tan B}{1 - \tan A \tan B} \]

4. **Substituting back**:
Now substituting \( \tan C \) back into our expression for \( \tan A + \tan B + \tan C \):
\[ \tan A + \tan B + \tan C = \tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

5. **Combine the terms**:
Let \( x = \tan A + \tan B \). Then we have:
\[ x - \frac{x}{1 - \tan A \tan B} \]
To combine these, we can find a common denominator:
\[ = \frac{x(1 - \tan A \tan B) - x}{1 - \tan A \tan B} \]
\[ = \frac{x - x + x \tan A \tan B}{1 - \tan A \tan B} \]
\[ = \frac{x \tan A \tan B}{1 - \tan A \tan B} \]

6. **Final equation**:
Therefore, we have:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
This implies:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]

### Conclusion:
Thus, the value of \( \tan A + \tan B + \tan C \) is equal to \( \tan A \tan B \tan C \).