若tan(3π+x)=2,cosx
若tan(3π+x)=2,cosx
tan(3x/2)-tan(x/2)-2sinx/(cosx+cos(2x))=?
证明tanx+1/cosx=tan(x/2+π/4)
证明1+sinx/cosx=tan(π/4+x/2)
求证:tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
证明:tan[3x/2]-tan[x/2]=2sinx/[cosx+cos2x]
证明:tan[3x/2]-tan[x/2]=2sinx/[cosx+cos2x]
证明tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
求证:tan(3x/2)-tan(x/2)=2sinx/(cosx+cos2x)
求证:tan(x-π/4)=(sinx-cosx)/(sinx+cosx)
已知sinx+cosx/sinx-cosx=3那么tan x
已知tan={根号2},x属于(π,3π/2),求sinx-cosx