Question

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

$x+z=2y$

${\tan }^{-1}x+{\tan }^{-1}z=2{\tan }^{-1}y$

${\tan }^{-1}\frac{x+z}{1-xz}=2{\tan }^{-1}y$

${\tan }^{-1}\frac{2y}{1-xz}={\tan }^{-1}\frac{2y}{1-y^2}$

$xz=y^2$

$x(x+2d)=(x+d{)}^2$

$x^2+2dx=x^2+2dx+d^2$

$d^2=0$

or,$d=0$

Hence, $x=y=z$