Question

If tan (A + B) = `sqrt3` and tan (A – B) = `1/sqrt3` ; 0° < A + B ≤ 90° ; A > B, find A and B.

Answer 1

tan (A + B) = √3 = tan 60° & tan (A – B) = 1/√3 = tan 30°

A + B = 60° …….(1)

A – B = 30° …….(2)

2A = 90° ⇒ A = 45°

adding (1) & (2)

A + B = 60

A – B = 30

Subtract equation (2) from (1)

A + B = 60

A – B = 30

2B = 30°

⇒ B = 15°. Ans.

Note: sin(A + B) = sin A cos B + cos A sin B

sin(A + B) ≠ sin A + sin B.