Q.Find the principal value of the following: $\sec^{-1} \left( \frac{2}{\sqrt{3}} \right)$

Concept: Inverse Secant Principal Value — For $\sec^{-1} x$, the principal value range is $[0, \pi]$, excluding $\frac{\pi}{2}$, and we choose the angle $\theta$ such that $\sec \theta = x$ and $\theta$ lies in that range.

1. Let $\theta = \sec^{-1} \left( \frac{2}{\sqrt{3}} \right)$. Then $\sec \theta = \frac{2}{\sqrt{3}}$, so $\cos \theta = \frac{\sqrt{3}}{2}$.
2. The angle whose cosine is $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$ (since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$). …

The inverse secant function, $\sec^{-1} x$, asks: "What angle $\theta$ has secant equal to $x$?" But because secant is not one-to-one over all real numbers, we restrict its domain to a specific interval to define a unique principal value. For $\sec^{-1}$, the standard principal range is $[0, \pi]$ excluding $\frac{\pi}{2}$ — that is, $\theta \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$. This ensures every input $x$ with $|x| \geq 1$ gives exactly one output.

The key trick: $\sec \theta = \frac{1}{\cos \theta}$, so solving $\sec^{-1}(a)$ is equivalent to solving $\cos \theta = \frac{1}{a}$, but you must then check that the resulting $\theta$ lies in the principal range of $\sec^{-1}$.

Let's apply this to $\sec^{-1}\left( \frac{2}{\sqrt{3}} \right)$.

1. Rewrite in terms of cosine. Let $\theta = \sec^{-1}\left( \frac{2}{\sqrt{3}} \right)$. Then by definition, $\sec \theta = \frac{2}{\sqrt{3}}$. Since $\sec \theta = \frac{1}{\cos \theta}$, we have:

2. Find all angles with that cosine. The equation $\cos \theta = \frac{\sqrt{3}}{2}$ is a standard trigonometric value. The reference angle is $\frac{\pi}{6}$ because $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$. Cosine is positive in the first and fourth quadrants, so the general solutions are:

Within one full cycle $[0, 2\pi)$, the angles are $\frac{\pi}{6}$ and $2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

3. Apply the principal range of $\sec^{-1}$. The principal value of $\sec^{-1}$ must lie in $[0, \pi]$ excluding $\frac{\pi}{2}$. Let's check each candidate:
   • $\frac{\pi}{6}$: This is in $[0, \frac{\pi}{2})$, so it is valid.
   • $\frac{11\pi}{6}$: This is greater than $\pi$ (since $\frac{11\pi}{6} \approx 3.67 > 3.14$), so it is not in the principal range. …

Method: Principal value of an inverse secant

Use this for $\sec^{-1}$ (and, by the same idea, $\csc^{-1}$) by converting to a cosine you already know.

Steps

Step 1: Convert secant to cosine

If $\theta = \sec^{-1}(k)$, then $\sec\theta = k$, so $\cos\theta = \frac{1}{k}$. This turns an unfamiliar secant into a standard cosine value.

Step 2: Recall the principal branch …

Common Mistakes

Mistake 1: Reading $\sec^{-1}$ as $\cos^{-1}$ of the same number

Why it's wrong: $\sec\theta$ is the reciprocal of $\cos\theta$, so you must first invert the argument: $\cos\theta = \frac{\sqrt{3}}{2}$, not $\frac{2}{\sqrt{3}}$. Correct approach: solve $\cos\theta = \frac{1}{k}$ before finding the angle.

Mistake 2: Forgetting that $\frac{\pi}{2}$ is excluded from the range …