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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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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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∵ tan(A + B) = √3 

∴ A + B = 60° …(1) 

∵ tan(A − B) = 1/√3 

∴ A − B = 30° …(2)

Adding (1) & (2), 

we get 2A = 90° 

⟹ A = 45°

Also (1) –(2), 

we get 2B = 30° 

⟹ B = 15°

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