Question

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

Answer 1

Answer: 0006

${\tan }^{-1}\left(\frac{1}{x}\right)+{\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}\left(\frac{1}{7}\right)\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{xy}}\right)={\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow \frac{x+y}{xy-1}=\frac{1}{7}\Rightarrow 7x+7y=xy-1\Rightarrow (7-y)x=-7y-1$
$\Rightarrow x=\frac{7y+1}{y-7}=\frac{7(y-7+7)+1}{y-7}\Rightarrow x=7+\frac{50}{y-7}$
Here, $y=8,9,12,17,32,57$ are satisfy. ]