The principal value branch of $ an^{-1} x$ is

(a) $\left[\frac{-\pi}{2}, \frac{\pi}{2} ight]$
(b) $[0, \pi]$
(c) $\left(\frac{-\pi}{2}, \frac{\pi}{2} ight)$
(d) $(0, \pi)$

Question

The principal value branch of $	an^{-1} x$ is

(a) $\left[\frac{-\pi}{2}, \frac{\pi}{2}ight]$
(b) $[0, \pi]$
(c) $\left(\frac{-\pi}{2}, \frac{\pi}{2}ight)$
(d) $(0, \pi)$

Accepted Answer

> [!TLDR]
> Tests the principal branch of $	an^{-1}x$: the open interval $\left(-	frac{\pi}{2},	frac{\pi}{2}ight)$. …

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

Q.The principal value branch of $\tan^{-1} x$ is