Q.Find the principal value of the following: $\operatorname{cosec}^{-1}(2)$

The key idea is that the principal value of $\operatorname{cosec}^{-1}(x)$ lies in the range $[-\pi/2, 0) \cup (0, \pi/2]$, excluding zero because cosecant is undefined there.

We want $\theta = \operatorname{cosec}^{-1}(2)$, so $\operatorname{cosec}(\theta) = 2$.

Since $\operatorname{cosec}(\theta) = \frac{1}{\sin(\theta)}$, this means $\sin(\theta) = \frac{1}{2}$.

The angles with $\sin(\theta) = 1/2$ in the principal range are $\theta = \pi/6$ (in $(0, \pi/2]$) and $\theta = 5\pi/6$ (outside the range). Only $\pi/6$ lies in the allowed interval.

The inverse cosecant function, $\operatorname{cosec}^{-1}(x)$, asks: "What angle $\theta$ (in the principal value range) has cosecant equal to $x$?" The key is the principal value range for $\operatorname{cosec}^{-1}$. Unlike the more familiar $\sin^{-1}$ which has range $[-\pi/2, \pi/2]$, the cosecant function is undefined at $\theta = 0$ (since $\sin 0 = 0$ and cosecant is $1/\sin\theta$). So the standard principal range for $\operatorname{cosec}^{-1}$ is:

This means we only consider angles in the first and fourth quadrants, excluding zero itself. Within this range, the cosecant function is one-to-one and covers all real numbers except those between $-1$ and $1$.

Now, $\operatorname{cosec}^{-1}(2)$ means we need $\theta$ such that $\operatorname{cosec}(\theta) = 2$. Since $\operatorname{cosec}(\theta) = \frac{1}{\sin\theta}$, this is equivalent to $\sin\theta = \frac{1}{2}$.

We know $\sin(\pi/6) = 1/2$. Is $\pi/6$ in the principal range? Yes — $\pi/6 \approx 0.523$ radians, which lies in $(0, \pi/2]$. So $\theta = \pi/6$ is a valid principal value.

Could there be another angle in the range? The other solution to $\sin\theta = 1/2$ in $[-\pi/2, 0)$ would be $\theta = -\pi/6$? Check: $\sin(-\pi/6) = -1/2$, not $1/2$. So no. The only candidate in the principal range is $\pi/6$.

Method: Evaluate $\operatorname{cosec}^{-1}$ by converting to sine

Steps

Step 1: Recall the range and domain.

The principal range of $\operatorname{cosec}^{-1}$ is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$, and the domain is $|x|\ge1$.

Step 2: Convert using the reciprocal.

$\operatorname{cosec}\theta=x\iff\sin\theta=\tfrac1x$, so $\operatorname{cosec}^{-1}(x)=\sin^{-1}\!\left(\tfrac1x\right)$ on the principal branch.

Step 3: Read off the standard angle: $\sin\theta=\tfrac12\Rightarrow\theta=\tfrac{\pi}{6}$, which lies in range.

Common Mistakes

Mistake 1: Writing $\operatorname{cosec}^{-1}(2)=\sin^{-1}(2)$.

Why it's wrong: it inverts the reciprocal incorrectly, and $\sin^{-1}(2)$ is undefined since $2\notin[-1,1]$. Correct approach: $\operatorname{cosec}^{-1}(2)=\sin^{-1}\frac12=\frac{\pi}{6}$.

Mistake 2: Forgetting the domain $|x|\ge1$.

Why it's wrong: $\operatorname{cosec}^{-1}$ rejects inputs in $(-1,1)$. Correct approach: confirm $|x|\ge1$ (here $2$) before evaluating.