`tanA +tanB + tanC = tanA tanB tanC` if

Options

A. A,B,C must be angles of a triangle
B. The sum of any two of A,B,C is equal to third
C. `A+B+C` must be integral multiple of `pi`.
D. `A+B+C` must be odd integral multiple of `pi`.

Answer

To solve the equation \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
We can rearrange this equation to group the terms:
\[ \tan A + \tan B = \tan A \tan B \tan C - \tan C \]

### Step 2: Factor out \( \tan C \)
Next, we can factor out \( \tan C \) from the right side:
\[ \tan A + \tan B = \tan C (\tan A \tan B - 1) \]

### Step 3: Divide both sides by \( \tan A \tan B - 1 \)
Assuming \( \tan A \tan B \neq 1 \), we can divide both sides by \( \tan A \tan B - 1 \):
\[ \frac{\tan A + \tan B}{\tan A \tan B - 1} = \tan C \]

### Step 4: Use the tangent addition formula
Recall the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
From our previous step, we can see that:
\[ \tan C = \frac{\tan A + \tan B}{\tan A \tan B - 1} \]
This implies:
\[ \tan C = -\tan(A + B) \quad \text{(if } A + B + C = 0\text{)} \]

### Step 5: Conclude the relationship
From the equation \( A + B + C = 0 \), we can conclude that:
\[ A + B = -C \]
Thus, the condition for the equality \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) to hold is:
\[ A + B + C = 0 \]

### Final Answer
The condition is \( A + B + C = 0 \).

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