Question

If \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \), where \( A + B + C = \pi \), then what is the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \)?

• A. 1
• B. 0
• C. 2
• D. Cannot be determined

Hint:

\textbf{Tip:} When \( A + B + C = \pi \), the identity \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) implies that the pairwise product sum is 1.

Answer 1

The Correct Option is A

Given the equations \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) and \( A + B + C = \pi \), we aim to find the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \).

Utilizing the identity \(\tan(A + B + C) = \tan \pi = 0\), we employ the tangent addition formula:

\(\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}\)

Substituting the given condition \(\tan A + \tan B + \tan C = \tan A \tan B \tan C\) into the formula yields:

\(0 = \frac{\tan A \tan B \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}\)

This simplifies to:

\(0 = \frac{0}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}\)

For this equation to hold, the numerator must be zero, which is satisfied. The denominator must be non-zero to avoid an undefined expression:

\(1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) eq 0\)

Therefore, it follows that \(\tan A \tan B + \tan B \tan C + \tan C \tan A = 1\).

The resultant value is **1**.