$\text{ If }\mathrm{x},\mathrm{y},\mathrm{z}\text{ are in A.P. and }{\tan }^{-1}\mathrm{x},{\tan }^{-1}\mathrm{y},{\tan }^{-1}\mathrm{z}\text{ are also in A.P. then }$

$$\begin{gathered} x,y,z are in A.P\Rightarrow 2y=x+z\to \left(1\right) \\ 2{Tan}^1\mathrm{y}={Tan}^{-1}\mathrm{x}+{Tan}^{-1}\mathrm{z} \\ \Rightarrow {\tan }^{-1}\left(\frac{2\mathrm{y}}{1-{\mathrm{y}}^2}\right)={\tan }^{-1}\left(\frac{\mathrm{x}+\mathrm{z}}{1-\mathrm{xz}}\right) \\ \Rightarrow \frac{2\mathrm{y}}{1-{\mathrm{y}}^2}=\frac{\mathrm{x}+\mathrm{z}}{1-\mathrm{xz}} \\ \mathrm{from}\left(1\right)\Rightarrow \frac{\mathrm{x}+\mathrm{z}}{1-{\mathrm{y}}^2}=\frac{\mathrm{x}+\mathrm{z}}{1-\mathrm{xz}} \\ \Rightarrow (\mathrm{x}+\mathrm{z})(\frac{1}{1-{\mathrm{y}}^2}-\frac{1}{1-\mathrm{xz}})=0 \\ \Rightarrow \mathrm{x}+\mathrm{z}=0 \mathrm{or} \frac{1}{1-{\mathrm{y}}^2}=\frac{1}{1-\mathrm{xz}} \\ \Rightarrow {\mathrm{y}}^2=\mathrm{xz} \\ \therefore \mathrm{x},\mathrm{y},\mathrm{z}\text{ are in G.P as well as in A.P } \\ \Rightarrow x=y=z \end{gathered}$$
$\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}$