China Mathematical Competition (Shaanxi)

Let $z=-x-\frac{\pi }{6}$. Then $z\in [\frac{\pi }{6},\frac{\pi }{4}]$, and $2z\in [\frac{\pi }{3},\frac{\pi }{2}]$. We have $$\tan \left(x+\frac{2\pi }{3}\right)=-\cot \left(x+\frac{6}{\pi }\right)=\cot z.$$ Then $$y=\cot z+\tan z+\cos z=\frac{2}{\sin 2z}+\cos z.$$ Since both $\frac{2}{\sin 2z}$ and $\cos z$ are monotonic decreasing in this case, so $y$ reaches the maximum at $z=\frac{\pi }{6}$, where $y_{\max }=\frac{2}{\sin \frac{\pi }{3}}+\cos \frac{\pi }{6}=\frac{4}{\sqrt{3}}+\frac{\sqrt{3}}{2}=\frac{11}{6}\sqrt{3}$.

Answer: C.