If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then

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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