由已知可得 tanα=tan[(α+π/4)-π/4]=[tan(α+π/4)-tan(π/4)] / [1+tan(α+π/4)tan(π/4)]=(1/5-1)/(1+1/5)= -2/3 ,所以 [sin(2α)-(sinα)^2] / [1-cos(2α)] =sinα(2cosα-sinα) / [2(sinα)^2]=(2cosα-si...