tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ
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tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ
要完整过程
要完整过程
tanαtanβ+1=(tanα-tanβ)/tan(α-β)
tanθtan2θ+1=(tan2θ-tanθ)/tan(2θ-θ)=(tan2θ-tanθ)/tanθ
tan2θtan3θ+1=(tan3θ-tan2θ)/tan(3θ-2θ)=(tan3θ-tan2θ)/tanθ
.
tan(nθ)tan(n+1)θ+1=[tan(n+1)θ-tan(nθ)]/tan[(n+1)θ-nθ]=[tan(n+1)θ-tan(nθ)]/tanθ
以上等式相加得
tanθtan2θ+tan2θtan3θ+……+tan(nθ)tan(n+1)θ+n
=(tan2θ-tanθ)/tanθ+(tan3θ-tan2θ)/tanθ+.+[tan(n+1)θ-tan(nθ)]/tanθ
=[tan(n+1)θ-tanθ]/tanθ
=tan(n+1)θ/tanθ-1
tanθtan2θ+tan2θtan3θ+……+tan(n-1)θtan(nθ)=tan(n+1)θ/tanθ-n-1
tanθtan2θ+1=(tan2θ-tanθ)/tan(2θ-θ)=(tan2θ-tanθ)/tanθ
tan2θtan3θ+1=(tan3θ-tan2θ)/tan(3θ-2θ)=(tan3θ-tan2θ)/tanθ
.
tan(nθ)tan(n+1)θ+1=[tan(n+1)θ-tan(nθ)]/tan[(n+1)θ-nθ]=[tan(n+1)θ-tan(nθ)]/tanθ
以上等式相加得
tanθtan2θ+tan2θtan3θ+……+tan(nθ)tan(n+1)θ+n
=(tan2θ-tanθ)/tanθ+(tan3θ-tan2θ)/tanθ+.+[tan(n+1)θ-tan(nθ)]/tanθ
=[tan(n+1)θ-tanθ]/tanθ
=tan(n+1)θ/tanθ-1
tanθtan2θ+tan2θtan3θ+……+tan(n-1)θtan(nθ)=tan(n+1)θ/tanθ-n-1
tanθ-tanθ分之1=-tan2θ分之2
证明:tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=tan(nα)/tanα-n
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