Find the principal value of sec⁻¹(-√2).

Let $\sec^{-1}(-\sqrt{2}) = \theta$, $\theta \in [0, \pi]$, $\theta \ne \frac{\pi}{2}$. $\sec\theta = -\sqrt{2} \Rightarrow \cos\theta = -\frac{1}{\sqrt{2}}$, so $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.