tan(α+π/4)=(tanα+tanπ/4)/(1-tanatanπ/4)=(tanα+1)/(1-tanα)=2
所以tanα=1/3
2cosα-sinα/cosα+3sinα=(2-sina/cosa)/(1+3sina/cosa)
=(2-tana)/(1+3tana)=5/6
tan(α+π/4)=(tanα+tanπ/4)/(1-tanatanπ/4)=(tanα+1)/(1-tanα)=2
所以tanα=1/3
2cosα-sinα/cosα+3sinα=(2-sina/cosa)/(1+3sina/cosa)
=(2-tana)/(1+3tana)=5/6