Q.Show that $\cot^{-1}\left(\dfrac{1}{\sqrt{x^2-1}}\right)=\sec^{-1}x$, $|x|>1$.

The three reciprocal trigonometric functions are inverted the same way — restrict to an interval where they are one-to-one, then define the inverse on that restricted range.

Inverse cosecant. $\text{cosec}\,x=\dfrac1{\sin x}$ has domain $\mathbb{R}\setminus\{n\pi\}$ and range $(-\infty,-1]\cup[1,\infty)$ (it never takes a value strictly between $-1$ and $1$). Restricting to $\left[-\tfrac{\pi}2,0\right)\cup\left(0,\tfrac{\pi}2\right]$ makes it a bijection onto that range, so $\text{cosec}^{-1}:(-\infty,-1]\cup[1,\infty)\to\left[-\tfrac{\pi}2,0\right)\cup\left(0,\tfrac{\pi}2\right]$ is defined by $\text{cosec}^{-1}x=y\iff\text{cosec}\,y=x$, $y$ in that restricted set. Its domain is compactly written $\mathbb{R}\setminus(-1,1)$ and its range $\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\}$.

Inverse secant. $\sec x=\dfrac1{\cos x}$ has domain $\mathbb{R}\setminus\left\{(2n+1)\tfrac{\pi}2\right\}$ and the same range $\mathbb{R}\setminus(-1,1)$. Restricting to $[0,\pi]\setminus\left\{\tfrac{\pi}2\right\}$ makes it a bijection onto that range, so $\sec^{-1}:\mathbb{R}\setminus(-1,1)\to[0,\pi]\setminus\left\{\tfrac{\pi}2\right\}$ is defined by $\sec^{-1}x=y\iff\sec y=x$, $y\in[0,\pi]\setminus\left\{\tfrac{\pi}2\right\}$.

Inverse cotangent. $\cot x=\dfrac1{\tan x}$ has domain $\mathbb{R}\setminus\{n\pi\}$ and range $\mathbb{R}$. Restricting to $(0,\pi)$ makes it a bijection onto $\mathbb{R}$, so $\cot^{-1}:\mathbb{R}\to(0,\pi)$ is defined by $\cot^{-1}x=y\iff\cot y=x$, $y\in(0,\pi)$. Unlike the other five, $\cot^{-1}x$ is defined for every real number with no gap. …