(sin2α-2cos^2α)/(1+tanα)＝(2sinαcosα-2cosα^2)/(1+tanα) =(2sinαcosα-2cosα^2)/[(1+tanα)(sinα^2+cosα^2)]=(2tanα-2)/[(1+tanα)(tanα^2+1)]因为tan(π/4+α)=-1/2＝(1+tanα)/(1-tanα),解出tanα...