Question

If ${\tan }^{-1}x$ , ${\tan }^{-1}y$ , ${\tan }^{-1}z$ are in A.P. and x, y, z are also in A.P. (y being not equal to 0, 1, or -1) , then :

• A. x, y, z are in G.P.
• B. ${(x-y)}^2+{(y-z)}^2+{(2z-x)}^2=0$
• C. $x=2y=z$
• D. none of these

Answer 1

Answer: (A)

x, y, z are in G.P.

Since $ta{n}^{-1}\left(x\right)$ , $ta{n}^{-1}\left(y\right)$ and $ta{n}^{-1}\left(z\right)$ are in A.P

$\Rightarrow 2ta{n}^{-1}\left(y\right)=ta{n}^{-1}\left(x\right)+ta{n}^{-1}\left(z\right)$

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

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

Now $2y=x+z$
Since x, y,z are in A.P
$\frac{2y}{1-y^2}=\frac{2y}{1-xz}$

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

$\Rightarrow xz=y^2$
Hence $x,y,z$ form a G.P.