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