If `tanalpha+tanbeta+tangamma=tanalphatanbetatangamma,then`

Options

A. `alpha, beta` must be angles of a triangle
B. the sum of any two of `alpha, beta,gamma` is equal to the third
C. `alpha+beta+gamma` must be an integral multiple of `pi`
D. none of these

Explanation

To solve the equation \( \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \), we can follow these steps:

### Step 1: Use the tangent addition formula
We know that the tangent of the sum of three angles can be expressed as:
\[ \tan(\alpha + \beta + \gamma) = \frac{\tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha)} \]

### Step 2: Substitute the given condition
Given that \( \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \), we can substitute this into the formula:
\[ \tan(\alpha + \beta + \gamma) = \frac{\tan \alpha \tan \beta \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha)} \]

### Step 3: Simplify the equation
The numerator simplifies to 0:
\[ \tan(\alpha + \beta + \gamma) = \frac{0}{1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha)} = 0 \]

### Step 4: Analyze the result
Since \( \tan(\alpha + \beta + \gamma) = 0 \), this implies that:
\[ \alpha + \beta + \gamma = n\pi \]
where \( n \) is any integer.

### Conclusion
Thus, we conclude that:
\[ \alpha + \beta + \gamma = n\pi \]
for some integer \( n \).