化简:cos^2θcotθ+sin^2θtanθ+2cosθsinθ
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/25 19:28:17
化简:cos^2θcotθ+sin^2θtanθ+2cosθsinθ
![化简:cos^2θcotθ+sin^2θtanθ+2cosθsinθ](/uploads/image/z/6127529-41-9.jpg?t=%E5%8C%96%E7%AE%80%3Acos%5E2%CE%B8cot%CE%B8%2Bsin%5E2%CE%B8tan%CE%B8%2B2cos%CE%B8sin%CE%B8)
原式=cos^3θ/sinθ+sin^3θ/cosθ+2cosθsinθ
=(cos^4θ+sin^4θ)/(sinθcosθ) +2cosθsinθ
=(cos^4θ+sin^4θ+2cos²θsin²θ)/(sinθcosθ)
=(cos²θ+sin²θ)²/(sinθcosθ)
=1/(sinθcosθ)
=(cos^4θ+sin^4θ)/(sinθcosθ) +2cosθsinθ
=(cos^4θ+sin^4θ+2cos²θsin²θ)/(sinθcosθ)
=(cos²θ+sin²θ)²/(sinθcosθ)
=1/(sinθcosθ)
化简:cos^2θcotθ+sin^2θtanθ+2cosθsinθ
证明1-2cos^2θ/tanθ-cotθ=sinθcosθ
求证:sin^2θtanθ+cos^2θcotθ+2sinθcosθ=tanθ+cotθ
化简【1+sinθ-cosθ/1+sinθ+cosθ】+cot(θ/2)
已知tanθ+cotθ=2,求sin^3θ-cos^3θ的值
sinθ-cosθ=1/2,则tanθ+cotθ=
已知sinθ+cosθ=√2,则tanθ+cotθ=?
sinΘ-cosΘ=1/2,求tanΘ+cotΘ
有道数学题帮下忙cot(sinθ)*tan(cosθ)
化简sin(θ-5π)•cot(π2-θ)•cos(8π-θ)tan(3π-θ)•tan(θ-32π)•sin(-θ-4
sin(θ -5π)/tan(3π-θ)*cot(π/2-θ)/tan(θ -3/2π)*cos(8π-θ)/sin(-
sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-(