Question

If \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) and \( A + B + C = \pi \), then which of the following is true?

• A. One of the angles is \( \frac{\pi}{2} \)
• B. All angles are equal
• C. \( \tan A = \tan B = \tan C \)
• D. \( A = B = \frac{\pi}{4} \)

Hint:

In trigonometric problems with \( A + B + C = \pi \), test for symmetry (e.g., equal angles) and use identities like \( \tan(\pi - x) = -\tan x \).

Answer 1

The Correct Option is B

- Given \( A + B + C = \pi \), we have \( C = \pi - (A + B) \). Using the identity \( \tan(\pi - x) = -\tan x \), we get: \[ \tan C = \tan(\pi - (A + B)) = -\tan(A + B) \]
Substitute into the given equation \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \): \[ \tan A + \tan B - \tan(A + B) = \tan A \tan B (-\tan(A + B)) \]
Use the identity \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). Let \( u = \tan A \), \( v = \tan B \), so \( \tan(A + B) = \frac{u + v}{1 - uv} \). The equation becomes: \[ u + v - \frac{u + v}{1 - uv} = uv \left(-\frac{u + v}{1 - uv}\right) \]
Simplify: \[ u + v - \frac{u + v}{1 - uv} + uv \frac{u + v}{1 - uv} = 0 \implies u + v = 0 \implies v = -u \]
Thus, \( \tan B = -\tan A \). Since \( \tan C = -\tan(A + B) \), and \( \tan(A + B) = 0 \) (as \( \tan A + \tan B = 0 \)), we have \( \tan C = 0 \), implying \( C = \pi \), which contradicts \( A + B + C = \pi \).
Instead, test the symmetric case: If \( A = B = C \), then \( A + B + C = 3A = \pi \implies A = \frac{\pi}{3} \). Check: \[ \tan \frac{\pi}{3} = \sqrt{3}, \quad \tan A + \tan B + \tan C = 3\sqrt{3}, \quad \tan A \tan B \tan C = (\sqrt{3})^3 = 3\sqrt{3} \]
This satisfies the equation. Hence, all angles are equal (\( A = B = C = \frac{\pi}{3} \)).