Let $u=\cot^{-1}x$. Solve $u^2-7u+10>0$, i.e., $(u-2)(u-5)>0$.

This gives $u<2$ or $u>5$.

Since $\cot^{-1}x$ is a decreasing function with range $(0,\pi)$:

• $u<2 \Rightarrow \cot^{-1}x<2 \Rightarrow x>\cot2$ (decreasing function reverses inequality)
• $u>5 \Rightarrow \cot^{-1}x>5 \Rightarrow x<\cot5$

Solution: $x\in(-\infty,\cot5)\cup(\cot2,\infty)$