tanα=1/2,tanβ=3,求tan(2α-β)的值

网友回答

tan2α=2tanα/[1-(tanα)^2]
=2*1/2/[1-(1/4)]
=1/(3/4)
=4/3tan(2α-β)
=(tan2α-tanβ)/[1+tan2αtanβ)
=[(4/3)-3]/[1+3*(4/3)]
=-(5/3)/5
=-1/3