已知tan^2A=2tan^B+1,求证sin^2B=2sin^2A-1

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已知tan^2A=2tan^B+1,求证sin^2B=2sin^2A-1

已知tan^2A=2tan^B+1,求证sin^2B=2sin^2A-1
已知tan^2A=2tan^B+1,求证sin^2B=2sin^2A-1

已知tan^2A=2tan^B+1,求证sin^2B=2sin^2A-1
即sin²a/cos²a=2sin²b/cos²b+1
sin²acos²b=2sin²bcos²a+cos²acos²b
sin²a(1-sin²b)=2sin²b(1-sin²a)+(1-sin²a)(1-sin²b)
sin²a-sin²asin²b=2sin²b-2sin²asin²b+1-sin²a-sin²b+sin²asin²b
2sin²a=sin²b+1
所以sin²b=2sin²a-1