Napier analogy and algebra in triangle.

Question

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.

Answer 1

$$\sum_{cyc}\frac{a-b}{a+b}=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}=$$ $$=\frac{\sum\limits_{cyc}c^2(a-b)}{\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}=-\prod_{cyc}\frac{a-b}{a+b}$$ and we are done!

Answer 2

Let $u = \tan \frac{B}{2}\tan \frac{C}{2}$, $v = \tan \frac{C}{2}\tan \frac{A}{2}$, $t = \tan \frac{A}{2}\tan \frac{B}{2}$. One has $u+v+t = 1$.

$x = \tan \frac{B-C}{2}\tan \frac{A}{2} = \frac{\tan\frac{B}{2}-\tan\frac{C}{2}}{1+\tan\frac{B}{2}\tan\frac{C}{2}}\tan \frac{A}{2} = \frac{t-v}{1+u}$.

$y = \frac{u-t}{1+v}$, $z= \frac{v-u}{1+t}$.

Thus, $x+y+z = \frac{t-v}{1+u} + \frac{u-t}{1+v} + \frac{v-u}{1+t} = \frac{\sum (t-v)(1+v+t+vt)}{(1+u)(1+v)(1+t)} = \frac{\sum vt(t-v)}{(1+u)(1+v)(1+t)} = -\frac{(v-u)(u-t)(t-v)}{(1+u)(1+v)(1+t)} = -xyz.$