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Let $x, y, z$ be positive real numbers such that $x, y, z$ are in GP and $\tan ^{-1} x, \tan ^{-1} y$ and $\tan ^{-1} z$ are in AP. Then which one of the following is correct?
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Verified Answer
The correct answer is:
$x=y=z$
$2 \tan ^{-1} \mathrm{y}=\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{z}$
$\left(\therefore \tan ^{-1} \mathrm{x}, \tan ^{-1} \mathrm{y}, \tan ^{-1} \mathrm{z}\right.$ are in A.P)
and $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in $\mathrm{G} \mathrm{P} \quad \therefore \quad \mathrm{y}^{2}=\mathrm{xz}$ ...(i)
$\Rightarrow \frac{2 y}{1-y^{2}}=\frac{x+z}{1-x z}$
$\Rightarrow 2 \mathrm{y}=\mathrm{x}+\mathrm{z}$
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in A.P.
Given $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are also in G.P
So, $x=y=z$
$\left(\therefore \tan ^{-1} \mathrm{x}, \tan ^{-1} \mathrm{y}, \tan ^{-1} \mathrm{z}\right.$ are in A.P)
and $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in $\mathrm{G} \mathrm{P} \quad \therefore \quad \mathrm{y}^{2}=\mathrm{xz}$ ...(i)
$\Rightarrow \frac{2 y}{1-y^{2}}=\frac{x+z}{1-x z}$
$\Rightarrow 2 \mathrm{y}=\mathrm{x}+\mathrm{z}$
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in A.P.
Given $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are also in G.P
So, $x=y=z$
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