因tanα=1/2,故tan2α=2tanα/(1-tan^2 α)=4/3
tan(2α+π/3)=(tan2α+tanπ/3)/(1-tan2αtanπ/3)
=[(4/3)+根号3]/[1-(4/3)(根号3)]
=- (48+25√3)/39
因tanα=1/2,故tan2α=2tanα/(1-tan^2 α)=4/3
tan(2α+π/3)=(tan2α+tanπ/3)/(1-tan2αtanπ/3)
=[(4/3)+根号3]/[1-(4/3)(根号3)]
=- (48+25√3)/39