Question

The principal value of $\sec^{-1}(-\frac{2}{\sqrt{3}})$ will be

• A. $-\frac{\pi}{2}$
• B. $-\frac{\pi}{6}$
• C. $-\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Answer 1

The principal value is $\frac{5\pi}{6}$, which is not among the given options.

Answer 2

Let

$$\sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) = \theta,$$

then

$$\sec\theta = -\frac{2}{\sqrt{3}} \quad \text{or} \quad \cos\theta = -\frac{\sqrt{3}}{2}.$$

The cosine function equals $-\frac{\sqrt{3}}{2}$ at

$$\theta = \frac{5\pi}{6} \quad (\text{or } 150^\circ)$$

in the principal range $[0,\pi]\setminus\{\pi/2\}$ for $\sec^{-1}(x)$.

Steps:

1. Expressed $\sec\theta$ in terms of $\cos\theta$.
2. Solved $\cos\theta = -\frac{\sqrt{3}}{2}$ within the principal range $[0, \pi],\, \theta \neq \frac{\pi}{2}$.
3. Obtained $\theta = \frac{5\pi}{6}$.