The function $f(x)={\tan }^{-1}x-x$ is monotonically decreasing in the set:

A function is monotonically decreasing if $f^{\prime }(x)<0$.Given $f(x)={\tan }^{-1}x-x$.Differentiating with respect to $x$:
$$f^{\prime }(x)=\frac{1}{1+x^2}-1$$
$$f^{\prime }(x)=\frac{1-(1+x^2)}{1+x^2}=\frac{-x^2}{1+x^2}$$
For any real $x\ne 0$, $x^2$ is positive, so $\frac{-x^2}{1+x^2}$ is always negative.If $x=0$, $f^{\prime }(x)=0$.Therefore, $f^{\prime }(x)<0$ for all $x\in \mathbb{R}-\{0\}$.