The $$+ ve$$  integral solution of $${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}{\text{ is}}$$

Solution :
$${\tan ^{ - 1}} + {\cos ^{ - 1}}\left( {\frac{y}{{1 + {y^2}}}} \right) = {\sin ^{ - 1}}\left( {\frac{3}{{\sqrt {10} }}} \right)$$
We will convert this equation in $${\tan ^{ - 1}}$$  from
$$\eqalign{ & \Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {\frac{1}{y}} \right) = {\tan ^{ - 1}}3 \cr & \Rightarrow {\tan ^{ - 1}}\left( {\frac{1}{y}} \right) = {\tan ^{ - 1}}3 - {\tan ^{ - 1}}x \cr & \Rightarrow {\tan ^{ - 1}}\left( {\frac{1}{y}} \right) = {\tan ^{ - 1}}\left( {\frac{{3 - x}}{{1 + 3x}}} \right) \cr & \Rightarrow \frac{1}{y} = \frac{{3 - x}}{{1 + 3x}} \cr & \Rightarrow y = \frac{{1 + 3x}}{{3 - x}} \cr} $$
As, we have to find positive integral values, so $$x$$ can have only two valus 1 and 2.
$$\eqalign{ & {\text{When }}x = 1,\,y = \frac{{1 + 3}}{2} \Rightarrow y = 2 \cr & {\text{When }}x = 2,\,y = \frac{{1 + 6}}{1} \Rightarrow y = 7 \cr} $$
So, these are two possible solutions for the given equation.