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

Solution :
Since, $$x, y, z$$   are in A.P.
⇒ $$2y = x + z$$
Also, we have, $$2\,{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( z \right)$$
$$\eqalign{ & \Rightarrow {\tan ^{ - 1}}\left( {\frac{{2y}}{{1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{x + z}}{{1 - xz}}} \right) \cr & \Rightarrow \,\,\frac{{x + z}}{{1 - {y^2}}} = \frac{{x + z}}{{1 - xz}}\,\,\left( {\because \,\,2y = x + z} \right) \cr & \Rightarrow \,\,{y^2} = xz\,\,\,{\text{or }}\,x + z = 0 \cr & \Rightarrow \,\,x = y = z \cr} $$