高一三角函数化简

化简：cos(x/2)/(√（1+sinx））+sin(x/2)/(√（1-sinx））（x属于（3π/2，2π）
1+sinx=1+2tan(x/2)/[1+tan(x/2)^2]=[tan(x/2)+1]^2/[1+tan(x/2)^2]
1-sinx=1-2tan(x/2)/[1+tan(x/2)^2]=[tan(x/2)-1]^2/[1+tan(x/2)^2]
-1<tan(x/2)<0 ,-1<cos(x/2)<0
cos(x/2)/(√（1+sinx））+sin(x/2)/(√（1-sinx））=√[1+tan(x/2)^2]*{cos(x/2)/[1-tan(x/2)]+sin(x/2)/[tan(x/2)+1]}
=√[1+tan(x/2)^2]*cos(x/2)*(cosx+sinx)/cosx=-(cosx+sinx)/cosx=-1-tanx

1+sinx=1+2tan(x/2)/[1+tan(x/2)^2]=[tan(x/2)+1]^2/[1+tan(x/2)^2]
1-sinx=1-2tan(x/2)/[1+tan(x/2)^2]=[tan(x/2)-1]^2/[1+tan(x/2)^2]
-1<tan(x/2)<0 ,-1<cos(x/2)<0
cos(x/2)/(√（1+sinx））+sin(x/2)/(√（1-sinx））=√[1+tan(x/2)^2]*{cos(x/2)/[1-tan(x/2)]+sin(x/2)/[tan(x/2)+1]}
=√[1+tan(x/2)^2]*cos(x/2)*(cosx+sinx)/cosx=-(cosx+sinx)/cosx=-1-tanx