In $\triangle \mathrm{ABC}$, given
$\begin{aligned}
& x=\tan \left(\frac{B-C}{2}\right) \tan \frac{A}{2} \\
& y=\tan \left(\frac{C-A}{2}\right) \tan \frac{B}{2}
\end{aligned}$
and $y=\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}$
Since, $\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \tan \frac{A}{2}$
$\Rightarrow \quad x=\frac{b-c}{b+c}$
Similarly, $y=\frac{c-a}{c+c}$ and $z=\frac{a-b}{a+b}$
Now, by componendo and dividendo
$\frac{1+x}{1-x}=\frac{b+c+b-c}{b+c-b+c}=\frac{b}{c}$
Similarly, $\frac{1+y}{1-y}=\frac{c}{a}$ and $\frac{1+z}{1-z}=\frac{a}{b}$
$\begin{aligned}
& \therefore \quad\left(\frac{1+x}{1-x}\right)\left(\frac{1+y}{1-y}\right)\left(\frac{1+z}{1-z}\right)=\frac{b}{c} \times \frac{c}{a} \times \frac{a}{b}=1 \\
\Rightarrow & (1+x)(1+y)(1+z) \\
= & (1-x)(1-y)(1-z) \\
\Rightarrow & 1+x+y+z+x y+y z+z x+x y z \\
= & 1-(x+y+z)+x y+y z+z x-x y z \\
\Rightarrow & x+y+z=-x y z
\end{aligned}$