If `A+C=B`, then tan A tan B tan C = ?

Options

A. `tanAtanB-tanC`
B. `tanB-tanC-tanA`
C. `tanA+tanB-tanC`
D. `-tanAtanB+tanC`

Answer

To solve the problem where \( A + C = B \) and we need to find the value of \( \tan A \tan B \tan C \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ A + C = B \]

### Step 2: Apply the tangent function
Taking the tangent of both sides, we get:
\[ \tan(A + C) = \tan B \]

### Step 3: Use the tangent addition formula
Using the formula for the tangent of a sum, we have:
\[ \tan(A + C) = \frac{\tan A + \tan C}{1 - \tan A \tan C} \]
Thus, we can write:
\[ \frac{\tan A + \tan C}{1 - \tan A \tan C} = \tan B \]

### Step 4: Rearranging the equation
Now, we can rearrange this equation to isolate \( \tan A + \tan C \):
\[ \tan A + \tan C = \tan B (1 - \tan A \tan C) \]
Expanding the right side gives:
\[ \tan A + \tan C = \tan B - \tan B \tan A \tan C \]

### Step 5: Rearranging terms
Now, we can bring all terms involving \( \tan A \) and \( \tan C \) to one side:
\[ \tan A + \tan C + \tan B \tan A \tan C = \tan B \]

### Step 6: Factor out \( \tan A \tan B \tan C \)
Rearranging gives us:
\[ \tan A \tan B \tan C = \tan B - \tan A - \tan C \]

### Step 7: Conclusion
From the equation we derived, we can conclude that:
\[ \tan A \tan B \tan C = \tan B - (\tan A + \tan C) \]
Since \( A + C = B \), we can substitute:
\[ \tan A \tan B \tan C = 0 \]

Thus, the final answer is:
\[ \tan A \tan B \tan C = 0 \]