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$\begin{aligned} \tan^{-1}x + \cos^{-1} \frac{y}{\sqrt{1+y^2}} & = \sin^{-1} \frac{3}{\sqrt{10}} \\ \Rightarrow \tan^{-1}x + \tan^{-1} \frac{1}{y} & = \tan^{-1} 3 \\ \tan \left( \tan^{-1}x + \tan^{-1} \frac{1}{y} \right) & = \tan \left( \tan^{-1} 3 \right) \\ \frac{x+\frac{1}{y}}{1-\frac{x}{y}} & = 3 \\ \frac{xy+1}{y-x} & = 3 \\ xy+ 1 & = 3y - 3x \\ xy + 3x - 3y + 1 & = 0 \\ (x-3)(y+3) + 10 & = 0 \\ \Rightarrow (x-3)(y+3) & = -10 \end{aligned}$

For $(x,y) \in \mathbb Z^+$ , $\Rightarrow \color{#D61F06}{(x-3)} \color{#3D99F6}{(y+3)} = \begin{cases} \color{#D61F06}{(-2)} \color{#3D99F6}{(5)} & \Rightarrow \color{#D61F06}{x = 1}, & \color{#3D99F6}{y = 2} \\ \color{#D61F06}{(-1)} \color{#3D99F6}{(10)} & \Rightarrow \color{#D61F06}{x = 2}, & \color{#3D99F6}{y = 7} \end{cases}$

$\Rightarrow \zeta = 2$ , $\beta = (1+2)(7-2) = 15$ and $\zeta + \beta = \boxed{17}$

The above equation can rewritten by few basics of inverse trig. As $\large\tan^{-1}x+\tan^{-1}\dfrac{1}{y}=\tan^{-1}3$ Solving this we get $y=\dfrac{3x+1}{3-x}$ and $x=\dfrac{3y-1}{3+y}$ Since $x,y$ are positive integers $x<3$ or $x=1,2$ the corresponding values of $y$ are $2,7$ .

Therefore the orders pairs are $\color{#D61F06}{(1,2)}$ and $\color{#D61F06}{(2,7)}$ .

We see that $\color{#3D99F6}{\zeta}=2$ and $\color{#3D99F6}{\beta}=15$ . Hence the answer is $\color{#EC7300}{15+2=\boxed{17}}$