3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0°B, find A and B.

tan (A + B) = √3 Since √3 = tan 60° Now substitute the degree value ⇒ tan (A + B) = tan 60° (A + B) = 60° … (i) The above equation is assumed as equation (i) tan (A – B) = 1/√3 Since 1/√3 = tan 30° Now substitute the degree value ⇒ tan (A – B) = tan 30° (A – B) = 30° … equation (ii) Now add the equation (i) and (ii), we get A + B + A – B = 60° + 30° Cancel the terms B 2A = 90° A= 45° Now, substitute the value of A in equation (i) to find the value of B 45° + B = 60° B = 60° – 45° B = 15° Therefore A = 45° and B = 15°