If tan(A+B) = √3 and tan(A-B) = 1/√3; 0 < A+B ≤ 90; A > B, find A and B.

tan(A+B) = √3, tan(A+B) = tan60° ⇒ A+B = 60° (i)

tan(A-B) = 1/√3
tan(A-B) = tan30° ⇒ A-B = 30° (ii)

From (i) and (ii), we get

2A = 90°
A = 45°

Substituting A = 45° in (i), we get
45° + B = 60°
B = 15°

Therefore, A = 45°, B = 15°