Solution :
Since, x, y and z are in AP
\(\therefore\) 2y = x + z
Also, \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are in AP
\(\therefore\) 2\(tan^{-1}y\) = \(tan^{-1}x\) + \(tan^{-1}z\)
\(\implies\) \(tan^{-1}({2y\over {1 – y^2}})\) = \(tan^{-1}({x + z\over {1 – xz}})\)
\(\implies\) \(x + z\over {1 – y^2}\) = \(x + z\over {1 – xz}\)
\(\implies\) \(y^2\) = xz
Since, x, y and z are in AP as well as in GP.
\(\therefore\) x = y = z
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