To solve this transcendental equations approximately :
Preivous: $\tanh(\alpha x) =\arctan(x)$, at $0<\alpha-1 \leq 1$ and at $\alpha\geq 1$.
Edit: $\tanh(\alpha x) =\arctan(x)$, at $0<\alpha-1 \ll 1$ and at $\alpha\gg 1$.
I need hint on how to start? I'm thinking to use Newton's Method of Iteration but don't know how?
On
We can have an extremely good estimate of the zero of function $$f(x)=\tanh (a x)-\tan ^{-1}(x)$$ making one iteration of Halley's mathod starting with $x_0=k$.
We have $$f(k)=\tanh (a k)-\tan ^{-1}(k)$$ $$f'(k)=a\, \text{sech}^2(a k)-\frac{1}{k^2+1}$$ $$f''(k)=\frac{2 k}{\left(k^2+1\right)^2}-2 a^2 \tanh (a k)\, \text{sech}^2(a k)$$ $$x_1=k+\frac{2\, f(k)\, f'(k)}{f(k) \,f''(k)-2 f'(k)^2}$$
After @Lutz Lehmann's comment, choose $k=\tan(1)$.
Some results
$$\left( \begin{array}{ccc} a & \text{estimation }& \text{solution} \\ 2 & 1.5432449040615292748 & 1.5432446279235773240 \\ 3 & 1.5568066556972442438 & 1.5568066556883083119 \\ 4 & 1.5573811263977578542 & 1.5573811263977572906 \\ 5 & 1.5574065442653326711 & 1.5574065442653326711 \end{array} \right)$$
For small values of $a$ $$\left( \begin{array}{ccc} a & \text{estimation }& \text{solution} \\ 1.1 & 1.1525726521622667697 & 1.1296349644856700163 \\ 1.2 & 1.3059331878896915304 & 1.3018024399310428716 \\ 1.3 & 1.3927625406966437033 & 1.3917949581555723633 \\ 1.4 & 1.4460683663131917454 & 1.4458089619371594747 \\ 1.5 & 1.4804871094885072890 & 1.4804115807485915832 \\ 1.6 & 1.5034696046136527262 & 1.5034463799965303017 \\ 1.7 & 1.5191789474986005266 & 1.5191715316939800960 \\ 1.8 & 1.5300991535713895614 & 1.5300967212675288608 \\ 1.9 & 1.5377849843609131701 & 1.5377841708190835133 \\ 2.0 & 1.5432449040615292748 & 1.5432446279235773240 \end{array} \right)$$
But, for small values of $a$, a Taylor expansion followed by a series reversion leads to $$x=1+t+\frac{ \left(4 a^2 \tanh ^3(a)-4 a^2 \tanh (a)+1\right)}{2 \left(-2 a+2 a \tanh ^2(a)+1\right)}t^2+O\left(t^3\right)$$ where $t=\frac{\pi -4 \tanh (a)}{4 a \text{sech}^2(a)-2}$
To solve $f(x)=0$ with$$f(x):=\tanh(\alpha x)-\arctan x\implies x-\frac{f}{f^\prime}=x-\frac{\tanh(\alpha x)-\arctan x}{\alpha\operatorname{sech}^2(\alpha x)-\tfrac{1}{1+x^2}}$$by Newton-Raphson, you just need a suitable choice of $x_0$. There's an obvious root at $x=0$ but, as we're not interested in that, we seek positive roots ($f$ is odd, so there's no point looking for negative ones; their locations follow from finding the positive ones). Indeed, for $\alpha>1$ we have a small-$x>0$ behaviour $f\sim(\alpha-1)x>0$, but $\lim_{x\to\infty}f(x)=1-\pi/2<0$, so some positive root exists.
We want $x_0$ to approximate this, so we don't converge to $x_0$ instead. For small $x$,$$\frac{f(x)}{x}\approx\alpha-1-\tfrac13(\alpha^3-1)x^2,$$which is $0$ at $x=x^\ast:=\frac{3}{\alpha^2+\alpha+1}$. @LutzLehmann, who has found this is too small a choice of $x_0$ for $\alpha-1\ll1$, has suggested $x_0=\tan1$, for which $\tanh(\alpha x)\approx1=\arctan x$. For $\alpha\gg1$, I expect $x_0=x^\ast$ to still be unhelpfully small.