<s>[INST] Napier analogy and algebra in triangle. 
If in a $\triangle{ABC}$, we define $x=\tan\frac{B-C}{2}\tan\frac{A}{2}$, $y=\tan\frac{C-A}{2}\tan\frac{B}{2}$ and $z=\tan\frac{A-B}{2}\tan\frac{C}{2}$, then show that $x+y+z=-xyz$.

My attempts:
By Napier analogy,
$x=\frac{b-c}{b+c},\ y=\frac{c-a}{c+a},\ z=\frac{a-b}{a+b}$
Then one can simply put these values in LHS, but that is cumbersome, I need to use some beautiful algebra, please help.
I just need to solve that algebra stuff, if such problem exists somewhere on this site then please comment with that link, I'll delete this, then.
 [/INST] $$\sum_{cyc}\frac{a-b}{a+b}=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}=$$
$$=\frac{\sum\limits_{cyc}c^2(a-b)}{\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}=-\prod_{cyc}\frac{a-b}{a+b}$$
and we are done!
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