若根号1-sinx/1+sinx=sinx-1/cosx,则x的取值范围是同题.选项为：A.2kπ≤x≤2kπ+π/2 B.2kπ+π/2

(sinx-1)/cosx>=0
分子小于等于0，分母小于0，所以x在二三象限
（题目最好打上括号）

√[(1-sinx)/(1+sinx)]=(sinx-1)/cosx,
将式中所有的1都换成sin²(x/2)+cos²(x/2),sinx都换成2sin(x/2)cos(x/2),cosx换成cos²(x/2)-sin²(x/2),得
√{[sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)]/[sin²(x/2)+2sin(x/2)cos(x/2)+cos²(x/2)]}
= -[sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)]/[cos²(x/2)-sin²(x/2)],
√{[sin(x/2)-cos(x/2)]²/[sin(x/2)+cos(x/2)]²}=
[sin(x/2)-cos(x/2)]²/{[sin(x/2)-cos(x/2)]*[sin(x/2)+cos(x/2)]},
| [sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)] |=[sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)],
上式实际上是|a|=a,说明a>0,所以
[sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)]>0
分子分母相除大于0,所以二者同号,所以二者相乘也大于0,即
[sin(x/2)-cos(x/2)]*[sin(x/2)+cos(x/2)]>0,变形
[sin(x/2)]²-[cos(x/2)]²>0
-cosx>0,（上面的等式是一个倍角公式）
cosx