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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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

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
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

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
A π＜x＜3π/2 π/2＜x/2＜3π/4 sinx