Question

For $0<x,y,z<1,$ if ${\tan }^{-1}x,{\tan }^{-1}y,{\tan }^{-1}z$ are in A.P. and $x,y,z$ are also in A.P., then

• A. $x,y,z$ are in G.P.
• B. $x,y,z$ are in H.P.
• C. $x=y=z$
• D. $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=0$

Answer 1

Answer: (A), (B), (C), (D)

$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=0$

${\tan }^{-1}x,{\tan }^{-1}y,{\tan }^{-1}z$ are in A.P.
$\Rightarrow 2{\tan }^{-1}y={\tan }^{-1}x+{\tan }^{-1}z$
$\Rightarrow 2{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+z}{1-xz}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{2y}{1-y^2}\right)={\tan }^{-1}\left(\frac{x+z}{1-xz}\right)$
$\Rightarrow \frac{2y}{1-y^2}=\frac{2y}{1-xz}[\because 2y=x+z]$
$\Rightarrow y^2=xz\cdots \left(1\right)$ $(\because y\ne 0)$
Hence, $x,y,z$ are in G.P.

Also, $x,y,z$ are in A.P.
$\Rightarrow x=y=z$
Hence, $x,y,z$ are in H.P.