Question

If $x$ and $y$ are positive integers satisfying $\tan ^{-1}\left(\frac{1}{x}\right)+\tan ^{-1}\left(\frac{1}{y}\right)=\tan ^{-1}\left(\frac{1}{7}\right)$, then the number of ordered pairs of $(x, y)$ is

Answer 1

Given equation is,
$ \tan ^{-1}\left(\frac{1}{x}\right)+\tan ^{-1}\left(\frac{1}{y}\right)=\tan ^{-1}\left(\frac{1}{7}\right) $
Using $\tan ^{-1} A+\tan ^{-1} B=\tan ^{-1}\left(\frac{A+B}{1-A B}\right)$, we get,
$ \begin{array}{l} \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{x y}}\right)=\tan ^{-1}\left(\frac{1}{7}\right) \\ \Rightarrow \frac{x+y}{x y-1}=\frac{1}{7} \\ \Rightarrow 7 x+7 y=x y-1 \\ \Rightarrow(y-7) x=7 y+1 \\ \Rightarrow x=\frac{7 y+1}{y-7}=7+\frac{50}{y-7} \end{array} $
Here, $y=8, 9,12,17, 32,57$ satisfy above equation.