Can there exist triangles ABC satisfying the following relation? Write yes or no giving reasons:
`tanA+tanB+tanC=0`

Answer

To determine if there can exist triangles \( ABC \) satisfying the relation \( \tan A + \tan B + \tan C = 0 \), we can analyze the situation step by step.

### Step-by-Step Solution:

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

2. **Rearranging the Angles**:
We can express \( A + B \) in terms of \( C \):
\[ A + B = 180^\circ - C \]

3. **Using the Tangent Addition Formula**:
We can apply the tangent addition formula to \( \tan(A + B) \):
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting \( A + B = 180^\circ - C \) gives:
\[ \tan(180^\circ - C) = -\tan C \]
Thus, we have:
\[ \frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C \]

4. **Setting Up the Equation**:
From the above, we can rewrite the equation:
\[ \tan A + \tan B = -\tan C (1 - \tan A \tan B) \]

5. **Substituting into the Given Condition**:
Now, we substitute \( \tan A + \tan B \) into the original condition:
\[ \tan A + \tan B + \tan C = 0 \]
This implies:
\[ \tan A + \tan B = -\tan C \]

6. **Combining the Results**:
From our previous steps, we can see that:
\[ -\tan C = -\tan C (1 - \tan A \tan B) \]
This leads to:
\[ -\tan C = -\tan C + \tan C \tan A \tan B \]
Simplifying this gives:
\[ 0 = \tan C \tan A \tan B \]

7. **Conclusion**:
The equation \( 0 = \tan C \tan A \tan B \) implies that at least one of the tangents must be zero. This means at least one of the angles \( A, B, \) or \( C \) must be \( 0^\circ \) or \( 180^\circ \), which is not possible in a triangle. Therefore, no triangle can satisfy the condition \( \tan A + \tan B + \tan C = 0 \).

### Final Answer:
**No, there cannot exist triangles ABC satisfying the relation \( \tan A + \tan B + \tan C = 0 \).**