如果tanx=2,那么sin^2(x)+sinx*cosx+cos^x是?
如果tanx=2,那么sin^2(x)+sinx*cosx+cos^x是?
求化简(sinx+tanx)/cos^2x+sin^2x+cosx
sin^2x*tanx+cos^2x*cotx+2sinx*cosx=tanx+cotx
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
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
1-2×sinx×cosx/cos∧2 x -sin∧2 x=1-tanx/1+tanx
已知(sinx+cosx)/(sinx-cosx)=3,求tanx,2sin²x+(sinx-cosx)&su
sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x
(2sinx*cosx)/sin^2x+cos^2x=(2tanx)/tan^2x+1 怎么推导的?
已知(1+tanx)/(1-tanx)=3 求(sin^2x+2sinx·cosx-cos^2x)/(sin^2x+2c