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) = `sqrt(3)` = tan 60° and tan (A – B) = `1/sqrt(3)` = tan 30°

A + B = 60°     ...(1)

A – B = 30°     ...(2)

2A = 90°

⇒ A = 45°

Adding (1) and (2)

A + B = 60°

A – B = 30°

Subtract equation (2) from (1)

A + B = 60°

A – B = 30°

2B = 30°

⇒ B = 15°

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

sin(A + B) ≠ sin A + sin B

Answer 2

Here, tan (A – B) = `1/sqrt(3)`

⇒ tan (A – B) = tan 30°          ...[∵ tan 30° = `1/sqrt(3)`]

⇒ (A – B) = 30°           ...(i)

Also, tan (A + B) = `sqrt(3)`

⇒ tan (A + B) = tan 60°           ...[∵ tan 60° = `sqrt(3)`]

⇒ A + B = 60^°                                 ...(ii)

Solving (i) and (ii), we get:

A = 45° and B = 15°