The problem asks us to find the exact values of $\sec(\frac{3\pi}{4})$ and $\cot(\frac{-11\pi}{6})$.

1. Problem Description

The problem asks us to find the exact values of $\sec(\frac{3\pi}{4})$ and $\cot(\frac{-11\pi}{6})$.

2. Solution Steps

10) We want to evaluate $\sec(\frac{3\pi}{4})$. Recall that $\sec(x) = \frac{1}{\cos(x)}$.

First, we determine $\cos(\frac{3\pi}{4})$.

The angle $\frac{3\pi}{4}$ is in the second quadrant. The reference angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$.

In the second quadrant, cosine is negative. Thus, $\cos(\frac{3\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$.

Then, $\sec(\frac{3\pi}{4}) = \frac{1}{\cos(\frac{3\pi}{4})} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\frac{2\sqrt{2}}{2} = -\sqrt{2}$.

11) We want to evaluate $\cot(\frac{-11\pi}{6})$. Recall that $\cot(x) = \frac{\cos(x)}{\sin(x)}$.

Also, $\cot(-x) = -\cot(x)$.

The angle $\frac{-11\pi}{6}$ is coterminal with $\frac{-11\pi}{6} + 2\pi = \frac{-11\pi}{6} + \frac{12\pi}{6} = \frac{\pi}{6}$.

Thus, $\cot(\frac{-11\pi}{6}) = \cot(\frac{\pi}{6})$.

We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

Therefore, $\cot(\frac{\pi}{6}) = \frac{\cos(\frac{\pi}{6})}{\sin(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \cdot \frac{2}{1} = \sqrt{3}$.

3. Final Answer

10) $-\sqrt{2}$

11) $\sqrt{3}$