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化简(sin^2x/sinx-cosx) -(sinx +cosx/tan^2x-1)
人气:408 ℃ 时间:2020-03-26 00:47:47
解答
[sin²x/(sinx-cosx]-[(sinx+cosx)/(tan²x-1)]=[sin²x/(sinx-cosx)]-[(sinx+cosx)cos²x/(sin²x-cos²x)]=[sin²x/(sinx-cosx)]-[cos²x/(sinx-cosx)]=(sin²x-cos²x)/...
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