已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值

tan(π/4+α)=sin(π/4+α)/cos(π/4+α)
=[sin(π/4)cosα+cos(π/4)sinα]/[cos(π/4)cosα-sin(π/4)sinα]
=[cosα+sinα]/[cosα-sinα]
=2
cosα+sinα=2cosα-2sinα
cosα=3sinα
tanα=1/3
1/(2sinαcosα+cos^2α)=1/[6(sinα)^2+9(sinα)^2]
=1/[15(sinα)^2]
2(sinα)^2=1-cos(2α)
cos(2α)=[1-(tanα)^2]/[1+(tanα)^2]
=[1-1/9]/[1+1/9]
=4/5
(sinα)^2=[1-cos(2α)]/2=1/10
1/(2sinαcosα+cos^2α)=1/[15(sinα)^2]=1/[15/10]=2/3