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

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1 Answers

Here, tan (A – B) = 1/√3 

⇒ tan (A – B) = tan 30 [∵ tan 30 = 1/√3 ] 

⇒ (A – B) = 30 …….(i) 

Also, tan (A + B) = √3

⇒ tan (A + B) = tan 60 [∵ tan 60 =√3] 

⇒ A + B = 60 …….(ii) 

Solving (i) and (ii), we get: 

A = 45 and B = 15

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