题目

已知tanx+1/tanx-1=3,求2sinx-3cosx/4sinx-9cosx；sin^2x-3sinxcosx+1

提问时间：2020-08-10

答案

(tanx+1)/(tanx-1)=3,tanx=2.∴(2sinx-3cosx)/(4sinx-9cosx) =(2tanx-3)/(4tanx-9) =(2×2-3)/(4×2-9) =-1.sin^2x-3sinxcosx+1 =(sin^2x-3sinxcosx)/(sin^2x+cos^2x)+1 =(tan^2x-3tanx)/(tan^2x+1)+1 =(4-3×2)/(4+...