$$\begin{align} (\cot\frac{1}{x})'=-\frac{1}{\tan^2\frac{1}{x}}\cdot\frac{1}{\cos^2\frac{1}{x}}\cdot(-\frac{1}{x^2})=\frac{1}{x^2\sin^2\frac{1}{x}}>0，x\in(1,+\infty) \end{align}\\$$ \begin{align} (\cot\frac{1}{x})'=-\frac{1}{\tan^2\frac{1}{x}}\cdot\frac{1}{\cos^2\frac{1}{x}}\cdot(-\frac{1}{x^2})=\frac{1}{x^2\sin^2\frac{1}{x}}>0，x\in(1,+\infty) \end{align}\\

解法有人给了，我就配个图吧