Question

If tan (A – B) = `1/sqrt(3) and tan (A + B) = sqrt(3), 0^0 ≤ (A + B) ≤ 90^0 and A > B`, then find A and B.

Answer 1

Here, tan (A – B) = `1/sqrt(3)`
⇒ tan (A – B) = tan `30^0   [∵ tan 30^0 = 1/sqrt(3)]`
⇒ (A – B) = `30^0`           …….(i)
Also, tan (A + B) = `sqrt(3)`

⇒ tan (A + B) = tan `60^0 [            ∵ tan 60^0 =sqrt(3)]`
⇒ A + B = 60⁰                                       …….(ii)
Solving (i) and (ii), we get:
A = 45⁰ and B = 15⁰