化简tanθ·tan2θ+tan2θ·tan3θ+.+tann·θ*tan(n+1)θ

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/05/06 09:07:52

∵1+tannθ*tan(n+1)θ=[cosnθcos(n+1)θ+sinnθsin(n+1)θ]/cosnθcos(n+1)θ
=cos[(n+1)θ-nθ]/cosnθcos(n+1)θ
=cosθ/cosnθcos(n+1)θ
=cotθ*sinθ/cosnθcos(n+1)θ
=cotθ*sin[(n+1)θ-nθ]/cosnθcos(n+1)θ
=cotθ[sin(n+1)θcosnθ/cosnθcos(n+1)θ-cos(n+1)θsinnθ/cosnθcos(n+1)θ]
=cotθ*[tan(n+1)θ-tannθ]
∴tannθ*tan(n+1)θ=cotθ*[tan(n+1)θ-tannθ]-1
故tanθ·tan2θ+tan2θ·tan3θ+.+tann·θ*tan(n+1)θ
=[cotθ(tan2θ-tanθ)-1]+[cotθ(tan3θ-tan2θ)-1]+.+cotθ[tan(n+1)θ-tannθ]-1
=cotθ[tan(n+1)θ-tanθ]-n
=cotθtan(n+1)θ-n-1
=tan(n+1)θ/tanθ-n-1

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