The problem is to evaluate the following expression: $\frac{sin(\frac{2\pi}{3}) + cos(\frac{\pi}{4})}{tan(\frac{\pi}{6}) - cot(\frac{\pi}{6})}$

1. Problem Description

The problem is to evaluate the following expression:

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

2. Solution Steps

First, we need to find the values of each trigonometric function:

$sin(\frac{2\pi}{3}) = sin(\pi - \frac{\pi}{3}) = sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$

$cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

$tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

$cot(\frac{\pi}{6}) = \frac{1}{tan(\frac{\pi}{6})} = \sqrt{3}$

Now substitute these values into the expression:

$\frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{3} - \sqrt{3}}$

Simplify the numerator:

$\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} = \frac{\sqrt{3} + \sqrt{2}}{2}$

Simplify the denominator:

$\frac{\sqrt{3}}{3} - \sqrt{3} = \frac{\sqrt{3}}{3} - \frac{3\sqrt{3}}{3} = \frac{\sqrt{3} - 3\sqrt{3}}{3} = \frac{-2\sqrt{3}}{3}$

Now we have:

$\frac{\frac{\sqrt{3} + \sqrt{2}}{2}}{\frac{-2\sqrt{3}}{3}} = \frac{\sqrt{3} + \sqrt{2}}{2} \cdot \frac{3}{-2\sqrt{3}} = \frac{3(\sqrt{3} + \sqrt{2})}{-4\sqrt{3}}$

Rationalize the denominator:

$\frac{3(\sqrt{3} + \sqrt{2})}{-4\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}(\sqrt{3} + \sqrt{2})}{-4(3)} = \frac{3(3 + \sqrt{6})}{-12} = \frac{3 + \sqrt{6}}{-4} = -\frac{3 + \sqrt{6}}{4}$

3. Final Answer

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