因为1=[sin(α/2)]^2+[cos(α/2)]^2, sinα=2sin(α/2)cos(α/2),
cosα=[cos(α/2)]^2-[sin(α/2)]^2
(1+sinα-cosα)/(1+sinα+cosα)
={[sin(α/2)+cos(α/2)]^2-[[cos(α/2)]^2-[sin(α/2)]^2]} / {[sin(α/2)+cos(α/2)]^2+[[cos(α/2)]^2-[sin(α/2)]^2]}
={[sin(α/2)]^2+sin(α/2)cos(α/2)} / {[cos(α/2)]^2+sin(α/2)cos(α/2)}
上下同除以[cos(α/2)]^2, 得到
={[tan(α/2)]^2+tan(α/2)} / [tan(α/2)+1]
=(5+√5)/(1+√5)
=√5