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

The correct option is D $x=y=z$

$x,y$ and $z$ are in $AP$

So, $2y=x+z$ ...(1)

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

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

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

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

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

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

From eqn. (1) $2y=x+z$

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

$(x+z)[\frac{1}{1-y^2}-\frac{1}{1-xz}]=0$

$x+z=0$ or $\frac{1}{1-y^2}-\frac{1}{1-xz}=0$

$1-y^2=1-xz$

$y^2=xz$

So, $x,y$ and $z$ are in $GP$
As by $x,y,z$ in $A.P.$ means
$2y=x+z$
upon squaring it gets down to
$4y^2=x^2+z^2+2xz=x^2+z^2+2y^2$ (from the above G.P. condition)
which resolves down to
$\Rightarrow x^2+z^2-2y^2=0\Rightarrow x^2+z^2-2xz=(x-z{)}^2=0$
thus, $x=z$
So,

If $x,y$ and $z$ are in $AP$ and $GP$ it means

$x=y=z$