sin2a=4/5 则tan^2a+cot^2a=
问题描述:
sin2a=4/5 则tan^2a+cot^2a=
答
sin2a=2sinacosa=4/5 => sinacosa=2/5tan^2a+cot^2a=(sin^4a+cos^4a)/(sin^2acos^2a)=[(sin^2a+cos^2a)^2-2sin^2acos^2a]/(sin^2acos^2a)=[1-2*(2/5)^2]/(2/5)^2=17/4