# If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

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

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Solution: tan (A + B) = √3

⇒ tan (A + B) = tan 60°

⇒ (A + B) = 60° ... (i)

tan (A – B) = 1/√3

⇒ tan (A - B) = tan 30°
⇒ (A - B) = 30° ... (ii)

Adding (i) and (ii), we get

A + B + A - B = 60° + 30°

2A = 90°

A= 45°

Putting the value of A in equation (i)

45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°

Thus, A = 45° and B = 15°

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