If `A+B+ C=(pi)/(2)` then what is the value of tan A tan B +tan B. tan C + tan C tan A?

Options

A. 0
B. 1
C. `-1`
D. tan A tan B tan C

Answer

To solve the problem, we need to find the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \) given that \( A + B + C = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Understand the Relationship**:
Since \( A + B + C = \frac{\pi}{2} \), we can express \( C \) in terms of \( A \) and \( B \):
\[ C = \frac{\pi}{2} - (A + B) \]

2. **Use the Tangent Addition Formula**:
We can apply the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Since \( C = \frac{\pi}{2} - (A + B) \), we know:
\[ \tan C = \tan\left(\frac{\pi}{2} - (A + B)\right) = \cot(A + B) \]
Therefore, we can write:
\[ \tan C = \frac{1}{\tan(A + B)} = \frac{1 - \tan A \tan B}{\tan A + \tan B} \]

3. **Substitute into the Expression**:
Now, we need to find \( \tan A \tan B + \tan B \tan C + \tan C \tan A \):
\[ \tan B \tan C = \tan B \cdot \frac{1 - \tan A \tan B}{\tan A + \tan B} \]
\[ \tan C \tan A = \tan A \cdot \frac{1 - \tan A \tan B}{\tan A + \tan B} \]

4. **Combine the Terms**:
We can combine these terms:
\[ \tan A \tan B + \tan B \cdot \frac{1 - \tan A \tan B}{\tan A + \tan B} + \tan A \cdot \frac{1 - \tan A \tan B}{\tan A + \tan B} \]

5. **Simplify**:
After simplification, we find that:
\[ \tan A \tan B + \tan B \tan C + \tan C \tan A = 1 \]

### Final Result:
Thus, the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \) is \( 1 \).