Given, ${\tan }^{-1}x, {\tan }^{-1}y$ and ${\tan }^{-1}z$ are in $\mathrm{A}.\mathrm{P}.$, so

$2{\tan }^{-1}y={\tan }^{-1}x+{\tan }^{-1}z$

$\Rightarrow {\tan }^{-1}y-{\tan }^{-1}x={\tan }^{-1}z-{\tan }^{-1}y$

$\frac{y-x}{1+xy}=\frac{z-y}{1+yz} ...\left(\mathrm{i}\right)$

Given $x, y, z$ are in $\mathrm{A}.\mathrm{P}.$

i.e., $2y=x+z\Rightarrow y-x=z-y$

Hence, from $\left(\mathrm{i}\right)$

$1+xy=1+yz$

$xy=yz$ either $y=0$ (rejected) or $x=z$

Hence, $x = y = z$.