Question

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

• A. $x=y=z$
• B. $n^{2}y=z$
• C. $z^2=xy$
• D. $y^2=\frac{1}{x2}$

Answer 1

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

Given $x,yz$ are in $AP$

So, $y=x=2-y$

$2y=x+2-\left(1\right)$

Similarly,

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

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

$\Rightarrow {\tan }^{-1}\left(\frac{2y}{1-y^2}\right)={\tan }^{-1}(x+\frac{z}{1-xz})\Rightarrow \frac{2y}{1-y^2}=x+\frac{z}{1-xz}\Rightarrow x+\frac{z}{1-y^2}=x+\frac{z}{1-xz}\Rightarrow \frac{1}{1-y^2}=\frac{1}{1-xz}\Rightarrow 1-xz=1-y^2\Rightarrow 1-xz=1-{\left\{\frac{x+z}{2}\right\}}^2\Rightarrow x^2+z^2+2xz=4xz\Rightarrow x^2+z^2+2xz-4xz=0\Rightarrow x^2+z^2-2xz=0\Rightarrow {(x-z)}^2=0$

$\Rightarrow x=z$ put in $\left(1\right)$

$\Rightarrow 2y=x+2\Rightarrow 2y=2x\Rightarrow y=xx=y=z$