已知sina=m,a∈(π/2,π),tan(π-b)=n(0<n<1),则tan(a-2b)=?
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已知sina=m,a∈(π/2,π),tan(π-b)=n(0<n<1),则tan(a-2b)=?
cosa=-√(1-m^2),
tana=-m/√(1-m^2),
tanb=-tan(π-b)=-n,
tan2b=2tanb/[1-(tanb)^2]=-2n/(1-n^2),
tan(a-2b)=(tana-tan2b)/(1+tanatan2b]
=[-m/√(1-m^2)]/{1+[-m/√(1-m^2)][-2n/(1-n^2)]}
=-2mn/[(1-n^2)√(1-m^2)+2mn].
tana=-m/√(1-m^2),
tanb=-tan(π-b)=-n,
tan2b=2tanb/[1-(tanb)^2]=-2n/(1-n^2),
tan(a-2b)=(tana-tan2b)/(1+tanatan2b]
=[-m/√(1-m^2)]/{1+[-m/√(1-m^2)][-2n/(1-n^2)]}
=-2mn/[(1-n^2)√(1-m^2)+2mn].
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