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 need to determine the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \). 

Using the identity: \(\tan(A + B + C) = \tan \pi = 0\), we have:

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

Substitute \(\tan A + \tan B + \tan C = \tan A \tan B \tan C\) into the equation:

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

This simplifies to:

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

Since the numerator is zero, the equation holds true regardless of the denominator (as long as it doesn't make the division undefined), implying:

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

Thus, \(\tan A \tan B + \tan B \tan C + \tan C \tan A = 1\).

Therefore, the value is 1.