If $x, y, \mathrm{z}$ are in Arithmetic Progression and $ an ^{-1} x, an ^{-1} y, an ^{-1} z$ are also in Arithmetic progression, where $x, \mathrm{z}\gt0$ and $x \mathrm{z} \lt 1, y \lt 1$, then

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If $x, y, \mathrm{z}$ are in Arithmetic Progression and $	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z$ are also in Arithmetic progression, where $x, \mathrm{z}\gt0$ and $x \mathrm{z} \lt 1, y \lt 1$, then

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Given, $x, y, z$ are in A.P. $	herefore \quad 2 y=x+z...(i)$ Also, $	an ^{-1} x, 	an ^{-1} y, 	an ^{-1} z$ are in A.P. $\begin{aligned} 	herefore \quad & 2 	an ^{-1} y=	an ^{-1} x+	an ^{-1} z \ & \Rightarrow 	an ^{-1}\left(\frac{2 y}{1-y^2}ight)=	an ^{-1}\left(\frac{x+z}{1-x z}ight) \ & \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-x z} \ & \Rightarrow \frac{2 y}{1-y^2}=\frac{2 y}{1-x z} ...[From(i)]\ & \Rightarrow 1-y^2=1-x z \ & \Rightarrow y^2=x z \end{aligned}$ $	herefore \quad x, y, \mathrm{z}$ are in G.P...(ii) From (i) and (ii), we get $x=y=\mathbf{z}$ $x=y=\mathrm{z}$

If $x, y, \mathrm{z}$ are in Arithmetic Progression and $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are also…

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