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

Question

If tan (A + B) = √3 and tan (A – B) = \(\frac{1}{\sqrt 3}\) 0° < A + B ≤ 90°; A > B. find A and B.

Answer 1

tan (A + B) = √3 ⇒ A + B = 60° ………….. (i) 

tan (A – B) = \(\frac{1}{\sqrt3}\) ⇒ A – B = 30° ………… (ii) 

From Eqn. (i) + Eqn.(ii), we have 

2A = 90 ⇒ = 45° 

From eqn. (i), A + B = 60° 

45° + B = 60° 

∴ B = 15