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

tan(a+π/4)=（tana+tanπ/4)/(1-tana*tanπ/4)=（tana+1)/(1-tana*1)=2
解得tana=1/3
lg(sina+2cosa)-lg(sina+3cosa)
=lg[(sina+2cosa)/(sina+3cosa)]
=lg[(sina+3cosa-cosa)/(sina+3cosa)]
=lg[1-cosa/(sina+3cosa)]
=lg[1-1/（tana+3）]
把tana=1/3代入,得原式=lg（7/10）=lg7-lg10=lg7-1