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