已知函数f(x)=(1-tanx)[1+(√2)sin(2x+π/4)],求函数定义域值域,最小正周期,单调递增区间
定义域：x≠kπ+π/2
f(x)=[(cosx-sinx)/cosx](1+sin2x+cos2x)=[(cosx-sinx)+(cosx-sinx)(sin2x+cos2x)]/cosx
=[(cosx-sinx)+sin2xcosx-sin2xsinx+cos2xcosx-cos2xsinx)]/cosx
=[(cosx-sinx)+(sin2xcosx-cos2xsinx)+(cos2xcosx-sin2xsinx)]/cosx
=[(cosx-sinx)+sin(2x-x)+cos(2x+x)]/cosx
=(cosx-sinx+sinx+cos3x)/cosx=(cosx+cos3x)/cosx=2cos2xcosx/cosx=2cos2x
故最小正周期T=π；
单调递增区间：由-π+2kπ≦2x≦2kπ,得单增区间为-π/2+kπ≦x≦kπ,k∈Z