Inverse Trigonometric equation.

$tan^{-1}(x-1)+tan^{-1}(x)+tan^{-1}(x+1)=tan^{-1}(3x)$

\Rightarrow\tan^{-1} (x-1 ) + \tan^{-1}(x+1) = \tan^{-1}(3x)-\tan^{-1}(x)

\Rightarrow \frac{tan(\tan^{-1} (x-1 )) + tan(\tan^{-1}(x+1))}{1-tan(\tan^{-1} (x-1 ))tan(\tan^{-1}(x+1))} = \frac{tan(\tan^{-1}(3x))-tan(\tan^{-1}(x)) }{1+tan(\tan^{-1}(3x))tan(\tan^{-1}(x))}

\Rightarrow \frac{x-1+x+1}{1-(x-1)(x+1)}=\frac{3x-x}{1+(3x)(x)}

\Rightarrow \frac{2x}{2-x^2}=\frac{2x}{1+3x^2}

\Rightarrow\boxed {x=0}

or when x≠0,

\Rightarrow 2-x^2=1+3x^2

\Rightarrow 4x^2=1

\Rightarrow \boxed{x=\pm 1/2}

tan ^-1 (x-1) + tan^-1(x+1) = tan^-13x -tan^-1x,

tan^-1(x-1+x +1)/1 - (x-1)(x+y)=tan^-1(3x-x)/1+3x² ,

2x /2-x² = 2x/1 +3x²,

x =0 or when x≠0 ,

4x² = 1,

x =±1/2

Yes, I am also getting the same answers.