麻烦证下这道数学题

设tan（x-y）=a,tan(y-z)=b,
则tan(z-x)=-tan（x-y+y-z）=(a+b)/(ab-1)
左边=a+b+(a+b)/(ab-1)
=(a+b)(1+1/(ab-1))
=(a+b)ab/(ab-1)
=ab(a+b)/(ab-1)
=右边
回答完毕.

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）

因为tan(a + b + c)
= [tan(a) + tan(b) + tan(c) - tan(a)*tan(b)*tan(c)] / [1 - tan(a)*tan(b) - tan(b)*tan(c) - tan(c)*tan(a)]
令a=x-y;b=y-z;c=z-x
则a+b+c=0故tan(a+b+c)=0
所以[tan(a) + tan(b) + tan(c) - tan(a)*tan(b)*tan(c)]