2.2.5 ** If $x \in\left[-\frac{5 \pi}{12},-\frac{\pi}{3}\right]$, then the maximum value of $y-\tan \left(x+\frac{2 \pi}{3}\right)-\tan \left(x+\frac{\pi}{6}\right)+$ $\cos \left(x+\frac{\pi}{6}\right)$ is ( ) .
(A) $\frac{12}{5} \sqrt{2}$
(B) $\frac{11}{6} \sqrt{2}$
(C) $\frac{11}{6} \sqrt{3}$
(D) $\frac{12}{5} \sqrt{3}$
Parse: First, transform $y$ into the algebraic sum of trigonometric functions of the same angle, we get
$$
\begin{aligned}
y & =-\cot \left(x+\frac{\pi}{6}\right)-\tan \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right) \\
& =\cot \left(-x-\frac{\pi}{6}\right)+\tan \left(-x-\frac{\pi}{6}\right)+\cos \left(-x-\frac{\pi}{6}\right).
\end{aligned}
$$

Let $z=-x-\frac{\pi}{6}$, then $z \in\left[\frac{\pi}{6}, \frac{\pi}{4}\right], 2 z=\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, and
$$
y=\cot z+\tan z+\cos z=\frac{2}{\sin 2 z}+\cos z,
$$

Clearly, $\frac{2}{\sin 2 z}$ and $\cos z$ are both decreasing, so the maximum value of $y$ is obtained when $z=\frac{\pi}{6}$, i.e., $y_{\max }=$ $\frac{2}{\sin \frac{\pi}{3}}+\cos \frac{\pi}{6}=\frac{4}{\sqrt{3}}+\frac{\sqrt{3}}{2}=\frac{11}{6} \sqrt{3}$. Therefore, the answer is C.