Question

已知tanɑ ∕tanɑ-1=-1,求sin²ɑ+sinɑcosɑ+2的值

Answer 1

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