1.

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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