若π<x<3π/2,则sqrt(tanx+sinx)+sqrt(tanx-sinx)=?A.2·sq

发布时间:2021-02-26 03:43:10

若π<x<3π/2,则sqrt(tanx+sinx)+sqrt(tanx-sinx)=?A.2·sqrt(tanx)·sin(x/2-π/4) B.2·sqrt(tanx)·sin(x/2+π/4)C.-2·sqrt(tanx)·sin(x/2-π/4) D.-2sqrt(tanx)`·sin(x/2+π/4)sqrt根号

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A π<x<3π/2 π/2<x/2<3π/4 sinx
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