Given that $\frac{\cos \frac{\pi}{6} - \tan \frac{\pi}{4}}{\sin \frac{\pi}{3} + \cos \frac{\pi}{3}} = \frac{\sqrt{3} + a}{\sqrt{3} + b}$, where $a$ and $b$ are integers.
Find the values of $a$ and $b$.

1. Substitute the exact values of the trigonometric functions

$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$
$\tan \frac{\pi}{4} = 1$
$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$
$\cos \frac{\pi}{3} = \frac{1}{2}$

$\text{LHS} = \frac{\frac{\sqrt{3}}{2} - 1}{\frac{\sqrt{3}}{2} + \frac{1}{2}}$

2. Simplify the fraction to match the required form

$\text{LHS} = \frac{\frac{\sqrt{3} - 2}{2}}{\frac{\sqrt{3} + 1}{2}}$
$\text{LHS} = \frac{\sqrt{3} - 2}{\sqrt{3} + 1}$

3. Compare with the given expression to find $a$ and $b$

Comparing $\frac{\sqrt{3} - 2}{\sqrt{3} + 1}$ with $\frac{\sqrt{3} + a}{\sqrt{3} + b}$:
$a = -2$
$b = 1$