如何证明 tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)

来源：百度知道编辑：UC知道时间：2024/05/22 13:10:20

您好！您的问题是这样的。用两倍角正切公式的变形，得tan(x-y)+tan(y-z)+tan(z-x)＝tan[(x-y)+(y-z)]*[1-tan(x-y)tan(y-z)]+tan(z-x)＝tan(x-z)*[1-tan(x-y)tan(y-z)]+tan(z-x)＝[tan(x-z)+tan(z-x)]-tan(x-z)tan(x-y)tan(y-z)=0+tan(x-y)tan(y-z)tan(z-x)＝tan(x-y)tan(y-z)tan(z-x)（此处用到正切函数是奇函数的性质）

如何证明 tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x) 证明:tan(x+y)tan(x-y)=tan②x-tan②y/1-tan②xtan②y tan(x+y)=0.4,tan(y-45°)=0.25,则tan(x+45°)=?要过程如何证明(X+Y)/100<2XY/(X+Y) y=tan(x-π/4) +tan(x π/4)的单调区间求证 [ tan(x+y) tan(x-y) ] = (sin^2 x - sin^2 y) / (cos^2 x - sin^2x) y=tan(x/2+pi/3)？ t=++x||--y；看不懂！ 3sinX=sin(2X+Y) 求证：tan(X+Y)=2tanX 2sinx+3sin(2y+x)=0,求5tan(x+y)+tany=