设(2cosx-sinx)(sin^2x+2cos^2x)=0,则(2cos^2x+sin^2x)/(1+tanx)等于
设(2cosx-sinx)(sin^2x+2cos^2x)=0,则(2cos^2x+sin^2x)/(1+tanx)等于
求化简(sinx+tanx)/cos^2x+sin^2x+cosx
1-2sinx cosx /COS^2X-SIN^2X =1-tanx/1+tanx 求证
提问数学难题求证:sin^2x*tanx+cos^2x/tanx+2sinx*cosx=tanx+1/tanx
sin^2x*tanx+cos^2x*cotx+2sinx*cosx=tanx+cotx
已知tanx=1/2,求sin x-cos x/2sinx+3cosx的值
(1+tanx)/(1-tanx)=3+2根号2,求(sin x+cosx)^2-(cos^3x)/sinx
如果tanx=2,那么sin^2(x)+sinx*cosx+cos^x是?
已知(1+tanx)/(1-tanx)=3 求(sin^2x+2sinx·cosx-cos^2x)/(sin^2x+2c
若|sinx|>|cosx|,则sin^2 x>cos^2 x
1-2×sinx×cosx/cos∧2 x -sin∧2 x=1-tanx/1+tanx
(2sinx*cosx)/sin^2x+cos^2x=(2tanx)/tan^2x+1 怎么推导的?