x∈(0,π/2)sin(x)+cos(x)=tan(x),x的范围?
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
证明(1-2sin x cos x )/(cos^2x-sin^2x)=(1-tan x)/(1+tan x)
已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
证明1-tan^2x/1+tan^2x=cos^2x-sin^2x
sin^3x-cos^3x≥cosx-sinx,求x的取值范围.x∈{0,2π}
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
求极限x趋于0时(1-cos(x/2))x / (tan x-sin x)
f(x)=|sin x+cos x+tan x+cot x+sec x+csc x|最小值
|cos x|=sin(3π/2-x),求角x的取值范围
已知sin x/2 -- 2cos x/2=0.(1)求tan x的值.(2)求 cos2x/(√2cos(π/4+x)
化简:1.【(sin^2)(-X-π) *cos(π+X)cosX】/【tan(2π+X) *(cos^3 (-X-π)
已知f(x)=(sin(π-x)*cos(2π-x))/cos(-π-x)*tan(π-x),则f(-31π/3)的值为