Q.Which of the following is the principal value branch of $\csc^{-1}x$?
(A) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(B) $[0,\pi]-\left\{\frac{\pi}{2}\right\}$
(C) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(D) $\left[\frac{-\pi}{2},\frac{\pi}{2}\right]-\{0\}$

The key idea is that the principal value branch of $\csc^{-1}x$ is chosen so that the function is one-to-one and covers all possible outputs except where $\csc \theta$ is undefined.

Step 1: Recall that $\csc \theta = 1/\sin \theta$, so $\csc^{-1}x$ is defined only where $\sin \theta \neq 0$. …

Why the principal value branch matters

Inverse trigonometric functions are defined by restricting the original trigonometric function to a domain where it is one‑to‑one. For $\csc x = \frac{1}{\sin x}$, the natural choice is to take the same interval used for $\sin^{-1}x$, but with a crucial adjustment: $\csc x$ is undefined wherever $\sin x = 0$, i.e., at $x = 0, \pm\pi, \pm2\pi,\dots$. So the principal branch must exclude those points.

The standard principal value branch for $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. For $\csc^{-1}x$, we use the same closed interval but remove the point where $\csc x$ blows up — that is, $x=0$. This gives $\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \setminus \{0\}$.

Let’s check each option carefully.

1. Option (A): $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

   This is an open interval. It excludes $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, but $\csc x$ is defined at both endpoints ($\csc(-\pi/2) = -1$, $\csc(\pi/2)=1$). More importantly, it still includes $0$, where $\csc x$ is undefined. So this cannot be the principal branch — it’s neither closed at the ends nor does it remove the problematic point.

2. Option (B): $[0,\pi] \setminus \left\{\frac{\pi}{2}\right\}$

   This interval runs from $0$ to $\pi$, excluding $\pi/2$. But $\csc x$ is undefined at $x=0$ and $x=\pi$ (since $\sin 0 = \sin \pi = 0$). Also, $\csc x$ takes both positive and negative values, and this interval only covers non‑negative outputs for $\csc^{-1}x$ — not the full range. So this is incorrect.

3. Option (C): $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

   This is identical to option (A). It has the same flaws: includes $0$, excludes endpoints unnecessarily. Not correct.

4. Option (D): $\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \setminus \{0\}$ …

Method: Principal branch of a reciprocal inverse function ($\csc^{-1}$, $\sec^{-1}$)

The reciprocal inverse functions inherit their branch from $\sin^{-1}$ or $\cos^{-1}$, with one point removed. Build the branch rather than recalling it blind.

Steps

Step 1: Start from the parent branch.

Since $\csc\theta=\dfrac{1}{\sin\theta}$, $\csc^{-1}$ uses the same closed interval as $\sin^{-1}$, namely $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. (For $\sec^{-1}$, start from $\cos^{-1}$'s branch $[0,\pi]$.)

Step 2: Remove the point where the reciprocal blows up.

A reciprocal is undefined wherever the denominator is zero. For $\csc^{-1}$, $\sin\theta=0$ at $\theta=0$, so delete $0$ from the interval. (For $\sec^{-1}$, delete $\tfrac{\pi}{2}$, where $\cos=0$.)

Step 3: Keep the endpoints — they are valid. …

Common Mistakes

Mistake 1: Choosing the open interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ to "avoid" $0$.

Why it's wrong: an open interval drops the endpoints $\pm\tfrac{\pi}{2}$, where $\csc$ is perfectly defined ($\csc(\pm\tfrac{\pi}{2})=\pm1$), yet still contains $0$ where $\csc$ blows up. Correct approach: keep the closed endpoints and remove only the single bad point: $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$. …