原式=(1+2cos²x-1)/[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]
=2cos²x/{[cos²(x/2)-sin²(x/2)]/sin(x/2)cos(x/2)}
=2cos²xsin(x/2)cos(x/2)/cos[2×(x/2)]
=cos²2x×sin[2×(x/2)]/cosx
=cosxsinx
=1/2×sin2x
原式=(1+2cos²x-1)/[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]
=2cos²x/{[cos²(x/2)-sin²(x/2)]/sin(x/2)cos(x/2)}
=2cos²xsin(x/2)cos(x/2)/cos[2×(x/2)]
=cos²2x×sin[2×(x/2)]/cosx
=cosxsinx
=1/2×sin2x