If tan(A + B) = √3 and tan(A − B) = 1/√3, where 0° B, then:

Question

If tan(A + B) = √3 and tan(A − B) = 1/√3, where 0° < A + B ≤ 90° and A > B, then:

Accepted Answer

A = 45°, B = 15°. A + B = 60° and A − B = 30°. Adding the two gives 2A = 90°, so A = 45°, and then B = 60° − 45° = 15°. Check: tan(45° + 15°) = tan60° = √3 and tan(45° − 15°) = tan30° = 1/√3.

If tan(A + B) = √3 and tan(A − B) = 1/√3, where 0° < A + B ≤ 90° and A > B, then:

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