Integral solutions to inverse trigonometric equation. The original question is this : I have to find number of ordered pairs of integral solutions to this :
$$\tan^{-1}{x} + \cos^{-1}{\frac{y}{\sqrt{1+y^2}}} = \sin^{-1}{\frac{3}{\sqrt{10}}} $$
I rearranged this as $\tan^{-1}(\frac{1}{y}) = tan^{-1}{3} - \tan^{-1}x $
And then $\frac{1}{y}=\frac{3-x}{1+3x}$.
Finally reduced it to $$(1+3x).(1-3y)=-10xy$$
The problem is $-10xy = -1.2.5.x.y$
So, do I really have to check $32$ cases? I know some of them can be discarded easily, but still, checking all will take a lot of time.
My question is, is there a different method to solve this? Or is there an easier way to check the cases?
Note : The time constraint of the question is ideally 3-4 minutes. This is a question for IIT-JEE which is an entrance exam for engineering in India.

$\bf{My\; Solution::}$ Given $$\displaystyle \arctan x + \arccos\left(\frac{y}{\sqrt{1+y^2}}\right) = \arcsin \left(\frac{3}{\sqrt{10}}\right)$$
Now We can Write $$\displaystyle \arccos\left(\frac{y}{\sqrt{1+y^2}}\right) = \arctan\left(\frac{1}{y}\right)$$ and $$\displaystyle \arcsin\left(\frac{3}{\sqrt{10}}\right)=\arctan (3)$$
So Our equation is convert into $$\displaystyle \arctan (x)+\arctan\left(\frac{1}{y}\right)=\arctan(3)$$
So we can write it as $$\displaystyle \arctan \left(\frac{x+\frac{1}{y}}{1-\frac{x}{y}}\right)=\arctan(3)\Rightarrow \frac{xy+1}{y-x}=3$$
So our expression is $$\displaystyle y=\frac{3x+1}{3-x} = 3\left(\frac{x+\frac{1}{3}}{3-x}\right).$$
Now If Given  $$\bf{x,y>0}$$ and $$\bf{x,y\in \mathbb{Z}^{+}}\;,$$ Then $$y=\left(\frac{x+\frac{1}{3}}{3-x}\right)>0$$
So We Get $\displaystyle -\frac{1}{3}<x<3$ and $x>0\;,$ Then we Get $0<x<3$
So Integer values of $x$ in This Interval are $x=1\;,2$
Now If $x=1\;,$ Then $\displaystyle y = \frac{3x+1}{3-x} = 2$ and If $x=2\;,$ Then $\displaystyle y=\frac{3x+1}{3-x} = 7$
So We get Only Two Positive Integer ordered pairs of $(x,y) = \left\{(1,2)\;\;,(2,7)\right\}$
