化简[（1+sinx）/cosx]*[sin2x/2cos^2(π/4-x/2)-tan(π/4-x/2)]

[（1+sinx）/cosx]*[sin2x/2cos^2(π/4-x/2)-tan(π/4-x/2)]
=[（1+sinx）/cosx]*{ sin2x/[1+cos(π/2 -x)] - [tan(π/4)-tan(x/2)]/[1+ tan(π/4)*tan(x/2)] }
=[(1+sinx）/cosx]*{ sin2x/(1+sinx) - [1- tan(x/2)]/[1+ tan(x/2)] }
=[(1+sinx）/cosx]*{ sin2x/[cos(x/2)+ sin(x/2)]² - [cos(x/2)- sin(x/2)]/[cos(x/2)+ sin(x/2)] }
=[(1+sinx）/cosx]*{ sin2x - [cos(x/2)- sin(x/2)]*[cos(x/2)+ sin(x/2)] }/[cos(x/2)+ sin(x/2)]²
=[(1+sinx）/cosx]*{ sin2x - [cos²(x/2)- sin²(x/2)]}/(1+sinx)
=(sin2x - cosx)/cosx
=2sinx-1
注：其中1+sinx=sin²(x/2)+2sin(x/2)*cos(x/2)+cos²(x/2)=[cos(x/2)+ sin(x/2)]²