函数y=sin2x+tanx/1-tan平方x,x∈[-π/8,π/8)的值域为?

y=sin2x+tanx/(1-tan²x)=sin2x+(1/2)tan2x∵x∈[-π/8,π/8],2x∈[-π/4,π/4],sin2x与tan2x均为增函数,∴y=si2x+(1/2)tan2x也是增函数,当x=-π/8时,y(min)=-√2/2-1/2,当x=π/8时,y(max)=√2/2+1/2,故该函数的...