<s>[INST] Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.
But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?
As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.
 [/INST] You do not need to consider cases.
Think first that you are looking for the zeros of function
$$f(x)=\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) -k$$ the first derivative
$$f'(x)=\frac{1}{1+(x+1)^2}+\frac{1}{1+x^2}+\frac{1}{1+(1-x)^2}=\frac {3 x^4+6 x^2+8 } {\left(x^2+1\right) \left(x^2-2 x+2\right) \left(x^2+2 x+2\right) }$$ is always positive; so, whatever is $k$, there is only one root.
$$k=\tan^{-1}(3)\implies f(0)=-\tan^{-1}(3)<0\quad \text{and}\quad f(1)=\frac{\pi }{4}-\cot ^{-1}(7)>0$$ So, the solution is $\in (0,1)$.
If you do not want to make anything complex, expand $f(x)$ as a series around $x=0$
$$f(x)=-\tan ^{-1}(3)+2 x-\frac{x^3}{6}+O\left(x^5\right)$$ Using the expansion to $O\left(x^3\right)$, the simplest estimate is
$$x_0 \sim \frac 12 \tan ^{-1}(3) =0.624523$$ while the exact solution is $5-\sqrt{19}=0.641101$.
If you already now about series reversion, using the expansion to $O\left(x^5\right)$ gives
$$x=\frac{1}{2} \left(f(x)+\tan ^{-1}(3)\right)+\frac{1}{96} \left(f(x)+\tan
   ^{-1}(3)\right)^3+O\left(\left(f(x)+\tan ^{-1}(3)\right)^5\right)$$ and since we want $f(x)=0$, then the estimate
$$x=\frac{1}{96} \tan ^{-1}(3) \left(48+\tan ^{-1}(3)^2\right)=0.644821$$
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