题目

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

提问时间：2020-08-02

答案

∵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