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

Answer 1

Answer: (D)

As $x,y,z$ are in A.P. $\Rightarrow 2y=x+z$ ...(i)
${\tan }^{-1}x,{\tan }^{-1}y$ and ${\tan }^{-1}z$ are also in A.P., then $2{\tan }^{-1}y={\tan }^{-1}x,+{\tan }^{-1}z$
$2{\tan }^{-1}y,+{\tan }^{-1}\left(\frac{x+z}{1-xz}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{2y}{1-y^2}\right)={\tan }^{-1}\left(\frac{x+z}{1-xz}\right)$
Thus $y^2=xz$... (ii)
From (i) and (ii), we get $x=y=z$.
Remark : $y\ne 0$ is implicit to make any of the choice correct.