已知cos2a+2sin^2a-sina=2/5,a含于(π/4,π),则tan(a+π/4)=

已知cos2a+2sin^2a-sina=2/5,a含于(π/4,π),则tan(a+π/4)=
已知cos2a+2sin^2a-sina=2/5,a含于(π/4,π),则tan(a+π/4)=

已知cos2a+2sin^2a-sina=2/5,a含于(π/4,π),则tan(a+π/4)=
cos2a+2sin^2a-sina=2/5
1-2sin^2a+2sin^2a-sina=2/5
sina=3/5
因为sin^2a+cos^2a=1,a含于(π/4,π)
所以cosa=±4/5
tana=±3/4
tan(a+π/4)=(1+tana)/(1-tana)
=7或1/7