已知α属于（π/2,π）,sinα=3/5,则tan(α+π/4)等于.

已知α属于（π/2,π）,sinα=3/5,则tan(α+π/4)等于.
已知α属于（π/2,π）,sinα=3/5,则tan(α+π/4)等于.

已知α属于（π/2,π）,sinα=3/5,则tan(α+π/4)等于.
因为a∈(π/2,π)且sina=3/5,则cosa=－4/5,所以tana=sina/cosa=－3/4
则：tan(a＋π/4)=[tana＋tan(π/4)]/[1－tanatan(π/4)]=(1＋tana)/(1－tana)=1/7

α属于（π/2，π），sinα=3/5，所以cosa= -4/5,tana= -3/4
tan(α+π/4)=(1-3/4)/(1-（-3/4）*1)=1/7

sinα=3/5 π/2<α<π
∴cosα=-4/5
tanα=sinα/cosα=-3/4
tan(α+π/4)=(tanα+tanπ/4)/(1-tanαtanπ/4)=(-3/4+1)/(1+3/4)=(1/4)/(7/4)=1/7