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

    化簡tanθ·tan2θ+tan2θ·tan3θ+.+tann·θ*tan(n+1)θ
    數(shù)學(xué)人氣:127 ℃時間:2020-04-03 12:50:01
    優(yōu)質(zhì)解答
    ∵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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