Question

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

Answer 1

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）