tan(x-y)=2,tan(x+y)=3 求tan2x/tan2y

tan2x=tan[(x+y)+(x-y)]=[tan(x+y)+tan(x-y)]/[1-tan(x+y)tan(x-y)]=(3+2)/(1-3×2)=-1
tan2y =tan[(x+y)-(x-y)]=[tan(x+y)-tan(x-y)]/[1+tan(x+y)tan(x-y)]=(3-2)/(1+3×2)=1/7
所以
tan2x/tan2y =-1/(1/7)=-7