tan(θ+π/4)=sin(2θ+π/2)/[1+cos(2θ+π/2)]=cos(2θ)/[1-sin(2θ)]=b/[1-a]ortan(θ+π/4)=[1-cos(2θ+π/2)]/sin(2θ+π/2)=[1+sin(2θ)]/cos(2θ)=[1+a]/b所以选D楼上两位解答都有问题:(1)jay_wei - 大魔法师 ...