We need to evaluate the expression: $\frac{sin(\frac{2\pi}{3}) + cos(\frac{\pi}{4})}{tan(\frac{\pi}{6}) - cot(\frac{\pi}{6})}$

1. Problem Description

We need to evaluate the expression:

$\frac{sin(\frac{2\pi}{3}) + cos(\frac{\pi}{4})}{tan(\frac{\pi}{6}) - cot(\frac{\pi}{6})}$

2. Solution Steps

First, we evaluate $sin(\frac{2\pi}{3})$. Since $\frac{2\pi}{3}$ is in the second quadrant, we have $sin(\frac{2\pi}{3}) = sin(\pi - \frac{\pi}{3}) = sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.

Next, we evaluate $cos(\frac{\pi}{4})$. We know that $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Then, we evaluate $tan(\frac{\pi}{6})$. We know that $tan(\frac{\pi}{6}) = \frac{sin(\frac{\pi}{6})}{cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

Finally, we evaluate $cot(\frac{\pi}{6})$. We know that $cot(\frac{\pi}{6}) = \frac{cos(\frac{\pi}{6})}{sin(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$.

Now, we substitute these values into the expression:

$\frac{sin(\frac{2\pi}{3}) + cos(\frac{\pi}{4})}{tan(\frac{\pi}{6}) - cot(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{3} - \sqrt{3}} = \frac{\frac{\sqrt{3} + \sqrt{2}}{2}}{\frac{\sqrt{3} - 3\sqrt{3}}{3}} = \frac{\frac{\sqrt{3} + \sqrt{2}}{2}}{\frac{-2\sqrt{3}}{3}} = \frac{\sqrt{3} + \sqrt{2}}{2} \cdot \frac{3}{-2\sqrt{3}} = \frac{3(\sqrt{3} + \sqrt{2})}{-4\sqrt{3}} = \frac{3(\sqrt{3} + \sqrt{2})(-4\sqrt{3})}{(-4\sqrt{3})(-4\sqrt{3})} = \frac{-12\sqrt{3}(\sqrt{3} + \sqrt{2})}{48} = \frac{-12(3 + \sqrt{6})}{48} = \frac{-(3 + \sqrt{6})}{4} = -\frac{3 + \sqrt{6}}{4}$

3. Final Answer

$-\frac{3 + \sqrt{6}}{4}$