<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] 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, $x+y+z = \frac{t-v}{1+u} + \frac{u-t}{1+v} + \frac{v-u}{1+t} = \frac{\sum (t-v)(1+v+t+vt)}{(1+u)(1+v)(1+t)} = \frac{\sum vt(t-v)}{(1+u)(1+v)(1+t)} = -\frac{(v-u)(u-t)(t-v)}{(1+u)(1+v)(1+t)} = -xyz.$
</s>