已知a∈（0,π/2）且tan(a+π/4)=3,则lg(sina+2cosa)-lg(3sina+cosa)的值为?

tan(a+π/4)=（tana+tanπ/4)/(1-tana*tanπ/4)=（tana+1)/(1-tana)=3
tana+1=3(1-tana)
4tana=2
解得tana=1/2
lg(sina+2cosa)-lg(3sina+cosa)
=lg[(sina+2cosa)/(3sina+cosa)]
=lg(sina/cosa+2)/(3sina/cosa+1)
=lg(tana+2)/(3tana+1)
=lg(1/2+2)/(3*1/2+1)
=lg(5/2)/(5/2)
=lg1
=0