Question

If $x,y,z$ are in A.P and ${\tan }^{-1}x,{\tan }^{-1}y\mathrm{and}{\tan }^{-1}z$ are also in A.P., then

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

Answer 1

The correct option is A

$x=y=z$

Explanation for the correct option.

$x,y,z$ are in A.P, so

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

${\tan }^{-1}x,{\tan }^{-1}y\mathrm{and}{\tan }^{-1}z$ are in A.P, so

$$\begin{gathered} \begin{array}{rcl}2{\tan }^{-1}y& =& {\tan }^{-1}x+{\tan }^{-1}z\\ & \Rightarrow & {\tan }^{-1}\left(\frac{2y}{1-y^2}\right)={\tan }^{-1}\left(\frac{x+z}{1-xz}\right)\left[by2{\tan }^{-1}\theta ={\tan }^{-1}\left(\frac{2\theta }{1-{\theta }^2}\right)and \\ {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\frac{A+B}{1-AB}\right]\\ \Rightarrow \frac{2y}{1-y^2}& =& \frac{2y}{1-xz}\left[from\left(1\right)\right]\\ \Rightarrow 1-xz& =& 1-y^2\\ \Rightarrow y^2& =& xz\end{array} \end{gathered}$$

This shows that $x,y,z$ are in A.P as well as in G.P.

Therefore, $x=y=z$

Hence, option A is correct.