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