已知∠A∈（0,π/2）且tan(A+π/4)=2,则lg(sinA+2cosA)-lg(sinA+3cosA)的值为?

题目

已知∠A∈（0,π/2）且tan(A+π/4)=2,则lg(sinA+2cosA)-lg(sinA+3cosA)的值为?

答案

tan(A+π/4)=2（tana+1）/(1-tana)=2tana+1=2-2tanatana=1/3lg(sinA+2cosA)-lg(sinA+3cosA)=lg[(sina+2cosa)/(sina+3cosa)]=lg[(tana+2)/(tana+3)]=lg[(1/3+2)/(1/3+3)]=lg(7/10)