Question

If `tan (A - B) = 1/sqrt(3)` and `tan (A + B) = sqrt(3)`, 0° < (A + B) ≤ 90° and A > B then find the values of A and B.

Answer 1

Given: `tan(A - B) = 1/sqrt(3), tan(A + B) = sqrt(3)`, 0° < (A + B) ≤ 90°, and A > B.

Step-wise calculation:

1. Since 0° < (A + B) ≤ 90° and `tan (A + B) = sqrt(3)`, the principal value is A + B = 60°   ...`(∵ tan 60^circ = sqrt(3))`

2. `tan(A - B) = 1/sqrt(3)` implies the principal acute value A – B = 30° (because `tan 30^circ = 1/sqrt(3)`) and A – B > 0 because A > B.

3. Solve the two linear equations:

A + B = 60°

A – B = 30°

Add them: 2A = 90°

⇒ A = 45°

Substitute: 45° + B = 60°

⇒ B = 15°