Question

If the non-zero numbers $x,y,z$ 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. $xy=yz$
• C. $x^2=yz$
• D. $z^2=xy$

Answer 1

Answer: (A)

Since $x,y,z$ are in $A.P.$
$\therefore 2y=x+z...\left(1\right)$
Since $ta{n}^{-1}x,ta{n}^{-1}y,ta{n}^{-1}z$ and in $A.P.$
$\therefore 2ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}z$
$ta{n}^{-1}\frac{2y}{1-y^2}=ta{n}^{-1}\frac{x+z}{1-xz}$
$\Rightarrow \frac{2y}{1-y^2}=\frac{x+z}{1-xz}$
$\therefore x,y,z$ are in $H.P.$
Also $y^2=xz$
$\Rightarrow x,y,z$ are in $G.P.$
$\therefore x=y=z$