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)

$x = y = z$

Answer 2

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 : $y \neq 0$ is implicit to make any of the choice correct.