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

Answer: 6

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