As $x, y, z$ are in A.P. $\Rightarrow \:\:\: 2y = x +z $ ...(i)
$\tan^{-1}x, \tan^{-1}y$ and $\tan^{-1}z$ are also in A.P., then $2\tan^{-1}y = \tan^{-1}x, + \tan^{-1}z$
$2\tan^{-1}y, + \tan^{-1} \left(\frac{x +z}{1 -xz} \right) $
$ \Rightarrow \:\:\:\tan^{-1} \left(\frac{2y}{1 - y^2} \right) = \tan^{-1} \left(\frac{x +z}{1 -xz} \right)$
Thus $y^2 = xz$... (ii)
From (i) and (ii), we get $x = y = z$.
Remark : $y \neq 0$ is implicit to make any of the choice correct.