已知tanα/tanα-1=-1,求sinα^2+sinαcosα+2的值
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tanα/(tanα-1)=-1
tanα=-(tanα-1)
tanα=-tanα+1
2tanα=1
tanα=1/2
sinα^2+sinαcosα+2
=(1-cos2α)/2+1/2*sin2α+2
=-1/2*cos2α+1/2*sin2α+5/2
=-1/2*[1-(tanα)^2]/[1+(tanα)^2]+1/2*2(tanα)^2/[1+(tanα)^2]+5/2
=-1/2*[1-(tanα)^2]/[1+(tanα)^2]+(tanα)^2/[1+(tanα)^2]+5/2
=-1/2*[1-(1/2)^2]/[1+(1/2)^2]+(1/2)^2/[1+(1/2)^2]+5/2
=-1/2*(3/4)/(5/4)+(1/4)/(5/4)+5/2
=-1/2*3/5+1/5+5/2
=-3/10+2/10+25/10
=24/10
=12/5
tanα=-(tanα-1)
tanα=-tanα+1
2tanα=1
tanα=1/2
sinα^2+sinαcosα+2
=(1-cos2α)/2+1/2*sin2α+2
=-1/2*cos2α+1/2*sin2α+5/2
=-1/2*[1-(tanα)^2]/[1+(tanα)^2]+1/2*2(tanα)^2/[1+(tanα)^2]+5/2
=-1/2*[1-(tanα)^2]/[1+(tanα)^2]+(tanα)^2/[1+(tanα)^2]+5/2
=-1/2*[1-(1/2)^2]/[1+(1/2)^2]+(1/2)^2/[1+(1/2)^2]+5/2
=-1/2*(3/4)/(5/4)+(1/4)/(5/4)+5/2
=-1/2*3/5+1/5+5/2
=-3/10+2/10+25/10
=24/10
=12/5
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