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

Answer

To solve the problem step by step, we will use the properties of arithmetic progression (A.P.) and the tangent function.

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since A, B, and C are in A.P., we can express A and C in terms of B:
\[ A = B - d \quad \text{and} \quad C = B + d \]
where \(d\) is the common difference.

2. **Substituting the Value of B**:
We are given that \(B = \frac{\pi}{4}\). Therefore, we can write:
\[ A = \frac{\pi}{4} - d \quad \text{and} \quad C = \frac{\pi}{4} + d \]

3. **Finding the Tangents**:
We need to find \(\tan A\), \(\tan B\), and \(\tan C\):
- For \(A\):
\[ \tan A = \tan\left(\frac{\pi}{4} - d\right) \]
- For \(B\):
\[ \tan B = \tan\left(\frac{\pi}{4}\right) = 1 \]
- For \(C\):
\[ \tan C = \tan\left(\frac{\pi}{4} + d\right) \]

4. **Using Tangent Addition and Subtraction Formulas**:
We can use the tangent subtraction and addition formulas:
- \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\)
- \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

Therefore:
\[ \tan A = \frac{\tan\left(\frac{\pi}{4}\right) - \tan d}{1 + \tan\left(\frac{\pi}{4}\right) \tan d} = \frac{1 - \tan d}{1 + \tan d} \]
\[ \tan C = \frac{\tan\left(\frac{\pi}{4}\right) + \tan d}{1 - \tan\left(\frac{\pi}{4}\right) \tan d} = \frac{1 + \tan d}{1 - \tan d} \]

5. **Calculating the Product**:
Now we need to find:
\[ \tan A \cdot \tan B \cdot \tan C = \left(\frac{1 - \tan d}{1 + \tan d}\right) \cdot 1 \cdot \left(\frac{1 + \tan d}{1 - \tan d}\right) \]

Simplifying this:
\[ = \frac{(1 - \tan d)(1 + \tan d)}{(1 + \tan d)(1 - \tan d)} = 1 \]

Thus, we conclude that:
\[ \tan A \cdot \tan B \cdot \tan C = 1 \]

### Final Answer:
\[ \tan A \tan B \tan C = 1 \]