Question

The number of integral pairs $(x,y)$ with $y>0$ satisfying ${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}$ is

Answer 1

${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}3$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}3-{\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}\left(\frac{3-x}{1+3x}\right)$
$\Rightarrow y=\frac{1+3x}{3-x}$
$y>0\Rightarrow x\in (-\frac{1}{3},3)$
Possible integral values of $x$ are $0,1,2$
But, for $x=0,y=\frac{1}{3}$ 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.