<s>[INST] $\tan^{-1}x$, $\tan^{-1}y$, $\tan^{-1}z$ are in arithmetic progression, as are $x$, $y$, $z$. Show ... 
$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:

*

*$x$, $y$, $z$ are in geometric progression

*$x$, $y$, $z$ are in harmonic progression.

*$x=y=z$

*$(x-y)^2 +(y-z)^2+(z-x)^2 =0$


My attempt:
$$\text{A}=\tan^{-1}x \qquad \text{B}=\tan^{-1}y \qquad \text{C}=\tan^{-1}z$$
$$x=\tan A \qquad y=\tan B \qquad  z=\tan C $$
$$x+z=2y$$
$$A+C=2B$$
$$\tan(A + B + C)=\frac{\tan A +\tan B +\tan C - \tan A\tan B\tan C }{1-\tan A\tan B -\tan B\tan C -\tan C\tan A}$$
$$\tan(3B)=\frac{x +y +z - xyz }{1-xy -yz -zx}$$
How do I continue from here?
 [/INST] HINT:
$$\tan^{-1}z-\tan^{-1}y=\tan^{-1}y-\tan^{-1}x$$
$$\implies\tan(\tan^{-1}z-\tan^{-1}y)=\tan(\tan^{-1}y-\tan^{-1}x)$$
$$\iff\dfrac{z-y}{1+yz}=\dfrac{y-x}{1+xy}$$
As $x,y,z$ are in A.P.,$y-x=z-y$
Case$\#1:$If If $y-x=z-y=0$
Case$\#2:$ If $y-x=z-y\ne0, 1+yz=1+xy\iff y(z-x)=0$
$\implies y=0$ or $z-x=0\iff z=x$
Can you take it from here?
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