If A,B,C are in A.P and `B=(pi)/4` then `tanAtanBtanC=`

Answer

To solve the problem, we need to find the value of \( \tan A \tan B \tan C \) given that \( A, B, C \) are in arithmetic progression (A.P.) and \( B = \frac{\pi}{4} \).

### Step-by-step Solution:

1. **Understanding the A.P.**:
Since \( A, B, C \) are in A.P., we can express \( A \) and \( C \) in terms of \( B \) and a common difference \( d \):
\[ A = B - d \quad \text{and} \quad C = B + d \]
Given \( B = \frac{\pi}{4} \), we can write:
\[ A = \frac{\pi}{4} - d \quad \text{and} \quad C = \frac{\pi}{4} + d \]

2. **Finding \( \tan A \), \( \tan B \), and \( \tan C \)**:
We need to calculate:
\[ \tan A = \tan\left(\frac{\pi}{4} - d\right), \quad \tan B = \tan\left(\frac{\pi}{4}\right), \quad \tan C = \tan\left(\frac{\pi}{4} + d\right) \]
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \).

3. **Using the tangent subtraction and addition formulas**:
We can use the tangent subtraction and addition formulas:
\[ \tan\left(\frac{\pi}{4} - d\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan d}{1 + \tan\left(\frac{\pi}{4}\right) \tan d} = \frac{1 - \tan d}{1 + \tan d} \]
\[ \tan\left(\frac{\pi}{4} + d\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan d}{1 - \tan\left(\frac{\pi}{4}\right) \tan d} = \frac{1 + \tan d}{1 - \tan d} \]

4. **Calculating the product \( \tan A \tan B \tan C \)**:
Now we can write:
\[ \tan A \tan B \tan C = \left(\frac{1 - \tan d}{1 + \tan d}\right) \cdot 1 \cdot \left(\frac{1 + \tan d}{1 - \tan d}\right) \]
Simplifying this expression:
\[ \tan A \tan B \tan C = \frac{(1 - \tan d)(1 + \tan d)}{(1 + \tan d)(1 - \tan d)} = 1 \]

### Final Answer:
Thus, the value of \( \tan A \tan B \tan C \) is:
\[ \boxed{1} \]