已知cot(45°+α)=2,则(1+sec2α-tan2α)/(1-sec2α+tan2α)=?
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已知cot(45°+α)=2,则(1+sec2α-tan2α)/(1-sec2α+tan2α)=?
cot(45°+α)=2
tan(45+α)=(1+tanα)/(1-tanα)=1/2
1-tanα=2+2tanα
3tanα=-1
tanα=-1/3
(1+sec2α-tan2α)/(1-sec2α+tan2α)
=(cos2α+1-sin2α)/(cos2α-1+sin2α)
=(2cos^2α-2sinαcosα)/(2sinαcosα-2sin^2α)
=(1-tanα)/(tanα-tanα^2)
=(1+1/3)/(-1/3-1/9)
=(4/3)/(-4/9)
=-1/3
tan(45+α)=(1+tanα)/(1-tanα)=1/2
1-tanα=2+2tanα
3tanα=-1
tanα=-1/3
(1+sec2α-tan2α)/(1-sec2α+tan2α)
=(cos2α+1-sin2α)/(cos2α-1+sin2α)
=(2cos^2α-2sinαcosα)/(2sinαcosα-2sin^2α)
=(1-tanα)/(tanα-tanα^2)
=(1+1/3)/(-1/3-1/9)
=(4/3)/(-4/9)
=-1/3
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