Question

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

• A. $x=y=z$
• B. $xy=yz$
• C. $x^2=yz$
• D. $z^2=xy$

Answer 1

Answer: (A), (B), (C), (D)

$z^2=xy$

Explanation for the correct options:

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

If $a,b,c$ are in AP then $2b=a+c$

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

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

$$\begin{gathered} \Rightarrow {\tan }^{-1}\left(\frac{2y}{1-y^2}\right)={\tan }^{-1}\left(\frac{x+z}{1-xz}\right);\left[\because 2{\tan }^{-1}A={\tan }^{-1}\left(\frac{2A}{1-A^2}\right),{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-\mathrm{AB}}\right)\right] \\ \Rightarrow \frac{2y}{1-y^2}=\frac{x+z}{1-xz} \\ \Rightarrow \frac{2y}{1-y^2}=\frac{2y}{1-xz}\left[\because 2y=x+z\right] \\ \Rightarrow 1-xz=1-y^2 \\ \Rightarrow xz=y^2-\left(ii\right) \end{gathered}$$

From $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} {\left(2y\right)}^2={\left(x+z\right)}^2 \\ \Rightarrow 4y^2=x^2+z^2+2xz \\ \Rightarrow 4\left(xz\right)=x^2+z^2+2xz \\ \Rightarrow x^2+z^2-2xz=0 \\ \Rightarrow {\left(x-z\right)}^2=0 \\ \Rightarrow x=z \end{gathered}$$

Since $y=\frac{x+z}{2},x=z$

$\therefore x=y=z$

So option A is correct,

For option B

$\because x=y=z$

$xy=yz$

So option B is correct

For option C

$\because x=y=z$

$x^2=yz$

So option C is correct

For option D

$\because x=y=z$

$z^2=xy$

So option D is correct

Hence option (A), option (B),option (C) and option (D) are correct.