tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ
要完整过程
人气:197 ℃ 时间:2020-05-13 16:09:48
解答
tanαtanβ+1=(tanα-tanβ)/tan(α-β)tanθtan2θ+1=(tan2θ-tanθ)/tan(2θ-θ)=(tan2θ-tanθ)/tanθtan2θtan3θ+1=(tan3θ-tan2θ)/tan(3θ-2θ)=(tan3θ-tan2θ)/tanθ.tan(nθ)tan(n+1)θ+1=[tan(n+1)θ...
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