The problem asks to evaluate the expression $\cos\frac{9\pi}{4} + \tan(-\frac{11\pi}{6})$.

1. Problem Description

The problem asks to evaluate the expression $\cos\frac{9\pi}{4} + \tan(-\frac{11\pi}{6})$.

2. Solution Steps

First, let's evaluate $\cos\frac{9\pi}{4}$.

Since $\cos(\theta + 2\pi n) = \cos(\theta)$ for any integer $n$, we have

$\cos\frac{9\pi}{4} = \cos(\frac{9\pi}{4} - 2\pi) = \cos(\frac{9\pi}{4} - \frac{8\pi}{4}) = \cos(\frac{\pi}{4})$.

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

Next, let's evaluate $\tan(-\frac{11\pi}{6})$.

Since $\tan(-\theta) = -\tan(\theta)$, we have $\tan(-\frac{11\pi}{6}) = -\tan(\frac{11\pi}{6})$.

Also, $\tan(\theta + \pi n) = \tan(\theta)$ for any integer $n$.

So, $-\tan(\frac{11\pi}{6}) = -\tan(\frac{11\pi}{6} - 2\pi) = -\tan(\frac{11\pi}{6} - \frac{12\pi}{6}) = -\tan(-\frac{\pi}{6}) = \tan(\frac{\pi}{6})$.

$\tan(\frac{\pi}{6}) = \frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

Therefore,

$\cos\frac{9\pi}{4} + \tan(-\frac{11\pi}{6}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} = \frac{3\sqrt{2} + 2\sqrt{3}}{6}$.

3. Final Answer

$\frac{3\sqrt{2} + 2\sqrt{3}}{6}$