Question

已知sina=m,a∈(π/2,π),tan(π-b)=n(0＜n＜1),则tan(a-2b)=?

Answer 1

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].