已知tan(π/4+A)=1/5,求 (sin2A-sin^A)/(1-COS2a)的值

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已知tan(π/4+A)=1/5,求 (sin2A-sin^A)/(1-COS2a)的值
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已知tan(π/4+A)=1/5,求 (sin2A-sin^A)/(1-COS2a)的值
已知tan(π/4+A)=1/5,求 (sin2A-sin^A)/(1-COS2a)的值

已知tan(π/4+A)=1/5,求 (sin2A-sin^A)/(1-COS2a)的值
令B = π / 4.-> tan(A + B) = (tanA - tanB )/(1+ tanA tanB) = 1 / 5 ->( tanA - 1 )/ (1+ tan A) =1 / 5
-> tan A = 3 / 2.(sin2A-sin^2A)/(1-COS2a) =( 2cosA sin A - sin^2(A) ) / ( 1 - (1 - 2sin^2(A) ) ) = cot (A) - 1 / 2 = 2 / 3 - 1 / 2 = 1 / 6