Maximizing $y=\tan(x+\frac{2\pi}{3})-\tan(x+\frac{\pi}{6})+\cos(x+\frac{\pi}{6})$ for $x\in[-\frac{5\pi}{12}, -\frac{\pi}{3}]$ 
Let $x \in [-\frac{5\pi}{12}, -\frac{\pi}{3}]$. Find the maximum value of $$y=\tan(x+\frac{2\pi}{3})-\tan(x+\frac{\pi}{6})+\cos(x+\frac{\pi}{6})$$

The above question is from the China Mathematical Competition from 2003, held for students in Shaanxi.
The solution that the testmakers made is provided below:
The solution (outline):
Let $z=-x-\frac{\pi}{6}$, then $\tan(x+\frac{2\pi}{3})=\cot(z)$ and then $y=\cot(z)+\tan(z)+\cos(z)=\frac{2}{\sin(2z)}+\cos(z)$. Then, both are monotonic decreasing, then $z=\frac{\pi}{6}$, and thus the minimum is $\frac{11}{6}\sqrt{3}$.

I just want to know any other methods to solve this question, especially because I find trigonometric inequalities and maxima and minima interesting.


The given proof is very clever and efficent.
As an alternative we have that
$$y=\frac{2}{\sin(2z)}+\cos(z)=\frac{1}{\sin(z)\cos(z)}+\cos(z)$$
which is equivalent to
$$y(t)=\frac{1}{t\sqrt{1-t^2}}+t \quad t=\left(\cos \left(\frac \pi 4\right),\cos \left(\frac \pi 6\right)\right)=\left(\frac{\sqrt 2}2,\frac{\sqrt 3}2\right)$$
with
$$\frac d{dt}\left(t\sqrt{1-t^2}\right)=\frac{1-2t^2}{\sqrt{1-t^2}}\le 0$$
$$0<t\sqrt{1-t^2}<1 \iff 0<t^2-t^4<1 \iff t^4-t^2+1 >0$$
therefore $y(t)$ is increasing and the maximum is reached for $z=\frac \pi 6$.
