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$x, y, z$ are in G.P. and $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P., then
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Verified Answer
The correct answer is:
$x=y=z$
$x, y, \mathrm{z}$ are in G.P.
$$
\Rightarrow y^2=x \mathrm{z}
$$
Also, $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P.
$$
\begin{aligned}
& \Rightarrow 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} \mathrm{z} \\
& \Rightarrow \tan ^{-1}\left(\frac{2 y}{1-y^2}\right)=\tan ^{-1}\left(\frac{x+\mathrm{z}}{1-x \mathrm{z}}\right) \\
& \Rightarrow \frac{2 y}{1-y^2}=\frac{x+\mathrm{z}}{1-x \mathrm{z}} \\
& \Rightarrow \frac{2 y}{1-x \mathrm{z}}=\frac{x+\mathrm{z}}{1-x \mathrm{z}}
\end{aligned}
$$
$$
\Rightarrow 2 y=x+\mathrm{z}
$$
$\Rightarrow x, y, \mathrm{z}$ are in A.P.
From (i) and (ii), we get
$$
x=y=\mathrm{Z}
$$
$$
\Rightarrow y^2=x \mathrm{z}
$$
Also, $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P.
$$
\begin{aligned}
& \Rightarrow 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} \mathrm{z} \\
& \Rightarrow \tan ^{-1}\left(\frac{2 y}{1-y^2}\right)=\tan ^{-1}\left(\frac{x+\mathrm{z}}{1-x \mathrm{z}}\right) \\
& \Rightarrow \frac{2 y}{1-y^2}=\frac{x+\mathrm{z}}{1-x \mathrm{z}} \\
& \Rightarrow \frac{2 y}{1-x \mathrm{z}}=\frac{x+\mathrm{z}}{1-x \mathrm{z}}
\end{aligned}
$$
$$
\Rightarrow 2 y=x+\mathrm{z}
$$
$\Rightarrow x, y, \mathrm{z}$ are in A.P.
From (i) and (ii), we get
$$
x=y=\mathrm{Z}
$$
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