已知tanɑ ∕tanɑ-1=-1,求sin²ɑ+sinɑcosɑ+2的值

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已知tanɑ ∕tanɑ-1=-1,求sin²ɑ+sinɑcosɑ+2的值

已知tanɑ ∕tanɑ-1=-1,求sin²ɑ+sinɑcosɑ+2的值
已知tanɑ ∕tanɑ-1=-1,求sin²ɑ+sinɑcosɑ+2的值

已知tanɑ ∕tanɑ-1=-1,求sin²ɑ+sinɑcosɑ+2的值
tanɑ ∕(tanɑ-1)=-1
tanα=1-tanα
2tanα=1
tanα=1/2
(sinα)^2=(tanα)^2/[1+(tanα)^2]=(1/4)/(1+1/4)=1/5
(sinɑ)^2+sinɑcosɑ+2
=(sinα)^2+sinα*sinα/tanα+2
=(sinα)^2+(sinα)^2/(1/2)+2
=3(sinα)^2+2
=3*1/5+2
=13/5=2.6

tana/(tana-1)=-1.===>2tana=1.tana=1/2.原式=[(tan²a+tana)/(tan²a+1)]+2=(3/5)+2=13/5.

tanɑ ∕(tanɑ-1)=-1
tanα=1-tanα
2tanα=1
tanα=1/2
(sinα)^2=(tanα)^2/[1+(tanα)^2]=(1/4)/(1+1/4)=1/5
(sinɑ)^2+sinɑcosɑ+2
=(sinα)^2+sinα*sinα/tanα+2
=(sinα)^2+(sinα)^2/(1/2)+2
=3(sinα)^2+2
=3*1/5+2
=13/5=2.6

由已知得,tana=1/2,cosa=2sina,sin²ɑ=1/5
sin²ɑ+sinɑcosɑ+2=3sin²ɑ+2=3*1/5+2=13/5