tan−1x+cos−1y√1+y2=sin−13√10
⇒tan−1x+tan−1(1y)=tan−13
⇒tan−1(1y)=tan−13−tan−1x
⇒tan−1(1y)=tan−1(3−x1+3x)
⇒y=1+3x3−x
y>0⇒x∈(−13,3)
Possible integral values of x are 0,1,2
But, for x=0,y=13 which is not an integer.
For x=1,y=2
For x=2,y=7
Therefore, there are two pairs (x,y) satisfying the given equation.