Question

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

• A. $x=y=z$
• B. $2x=3y=6z$
• C. $6x=3y=2z$
• D. $6x=4y=3z$

Answer 1

The correct option is A $x=y=z$

$x,y,z$ are in A.P.
So, $2y=x+z$
${\tan }^{-1}x,{\tan }^{-1}y,{\tan }^{-1}z$ are also in A.P.
So, $2{\tan }^{-1}y={\tan }^{-1}x+{\tan }^{-1}z$
$\Rightarrow {\tan }^{-1}\left(\frac{2y}{1-y^2}\right)={\tan }^{-1}\left(\frac{x+z}{1-xz}\right)$
$\Rightarrow \frac{2y}{1-y^2}=\frac{x+z}{1-xz}$
$\Rightarrow 2y(y^2-xz)=0$
$\Rightarrow y=0$ and $y^2=xz$
$\therefore x,y,x$ are in A.P. and G.P. both
Hence, $x=y=z$