Question

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

• A. $x=y=z$
• B. $x=y=-z$
• C. $x=1,y=2,z=3$
• D. $x=2,y=4,z=6$
• E. $x=2,y=3z$

Answer 1

The correct option is A

$x=y=z$

Explanation for the correct answer:

$x,y$ and $z$ are in arithmetic progression

$\Rightarrow$$2y=x+z$ $...\left(i\right)$

${\tan }^{-1}x,{\tan }^{-1}y$and ${\tan }^{-1}z$ are in an arithmetic progression

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

$\Rightarrow$$2{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+z}{1-xz}\right)$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-\mathrm{AB}}\right)\right]$

Taking tangent on both sides we get

$\Rightarrow$ $\tan \left(2{\tan }^{-1}y\right)=\tan \left({\tan }^{-1}\left(\frac{x+z}{1-xz}\right)\right)$

$\Rightarrow$$\frac{2\tan \left({\tan }^{-1}y\right)}{1-{\left[\tan \left({\tan }^{-1}y\right)\right]}^2}=\left(\frac{x+z}{1-xz}\right)$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \tan 2\theta =\frac{2\mathrm{tan\theta }}{1-{\tan }^2\theta },\mathrm{where}\theta ={\tan }^{-1}y\right]$

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

$\Rightarrow$ $\frac{x+z}{1-y^2}=\frac{x+z}{1-xz}$ …[From(i)]

$\Rightarrow$ $1-y^2=1-xz$

$\Rightarrow$ $y^2=xz$

$\Rightarrow$ $y$ is the geometric mean of $x$ and $z$. Hence, $x,y$ and $z$ are in a geometric progression

As $x,y$ and $z$ are in an arithmetic as well as geometric progression , the only possible relation between them is $x=y=z$.

Hence, option (A) is the correct answer.