Cotangents

To solve this problem we can start by substituting the first equation into the second and then solve for the value of $y$ . Substituting the two equations gives us $\cot (\frac{\pi}{4} + y) + \cot y = 2$ . We can now solve this equation from $[0, 2 \pi)$ :

$\cot (\frac{\pi}{4} + y) + \cot y = 2 \Rightarrow \frac{1}{\tan (\frac{\pi}{4} + y)} + \cot y = 2 \Rightarrow \frac{1 - \tan \frac{\pi}{4} \tan y}{\tan \frac{\pi}{4} + tan y} + \cot y = 2 \Rightarrow \frac{1 - \tan y}{1 + \tan y} + \cot y = 2 \Rightarrow \frac{1 - \tan y + \cot y + \cot y \tan y}{1 + \tan y} = 2 \Rightarrow \frac{2 - \tan y + \cot y}{1 + \tan y} - 2 = 0 \Rightarrow$ $\frac{2 - \tan y + \cot y - 2 - 2 \tan y}{1 + \tan y} = 0 \Rightarrow \frac{\cot y - 3 \tan y}{1 + \tan y} = 0 \Rightarrow \cot y - 3 \tan y = 0 \Rightarrow \frac{1 - 3 \tan ^2 y}{\tan y} = 0 \Rightarrow 1 - 3 \tan ^2 y = 0 \Rightarrow \tan y = \pm \frac{1}{\sqrt{3}} \Rightarrow y = {\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}}$

Once we have a value for $y$ we can take the smallest value and solve for $x = \frac{5 \pi}{12}$ , and then find $x + y = \frac{7}{12}$ , so $a + b = \boxed{19}$