zwwsr
幼苗
共回答了14个问题采纳率:85.7% 举报
因为tanx=2,所以cosx不等于0,
(1)(sinx-3cosx)/(sinx-cosx)
=(tanx-3)/(tanx-1) (上下除以cosx)
=(2-3)/(2-1)
=-1;
(2)[(sinx)^2-sinxcosx-2(cosx)^2]/[(cosx)^2-3sinxcosx-(sinx)^2]
=[(tanx)^2-tanx-2]/[1-3tanx-(tanx)^2] (上下除以(cosx)^2)
=(4-2-2)/(1-6-4)
=0;
(3)sinxcosx
=(sinxcosx)/[(sinx)^2+(cosx)^2]
=tanx/[(tanx)^2+1) (上下除以(cosx)^2)
=2/(4+1)
=2/5.
1年前
4