$y=cot^22x+tanx+cotx=(\dfrac{1-tan^2x}{2tanx})^2+tanx+\dfrac{1}{tanx}$
$=\dfrac{1}{4tan^2x}+\dfrac{tan^2x}{4}-\dfrac{1}{2}+\dfrac{1}{tanx}+tanx$
Đặt $\dfrac{1}{tanx}+tanx=t$
\Rightarrow $y=\dfrac{1}{4}(t^2-2)-\dfrac{1}{2}+t$
$=(\dfrac{1}{2}t+1)^2-2$
Do $x$ thuộc $[\dfrac{\pi}{12};\dfrac{\pi}{8}]$ nên $tanx$ đồng biến
\Rightarrow $2-\sqrt{3} \le tanx \le \sqrt{2}-1$
\Rightarrow ...