State the range of $\tan^{-1} p$. Given that $\cos^{-1} k = x$, where $0 \le x \le \frac{\pi}{2}$, find the principal value of $\cos^{-1}(-k)$.

1. State the principal value range of the inverse tangent function

The range of $\tan^{-1} p$ is $-\frac{\pi}{2} < \tan^{-1} p < \frac{\pi}{2}$.

1. Express $\cos^{-1}(-k)$ in terms of $x$

Let $y = \cos^{-1}(-k)$. Then $\cos y = -k$.
Since $\cos^{-1} k = x$, we have $\cos x = k$.
Therefore, $\cos y = -\cos x$.
Since the principal value range for $\cos^{-1}$ is $[0, \pi]$ and $0 \le x \le \frac{\pi}{2}$, the angle $y$ in the second quadrant is $\pi - x$.
Thus, $\cos^{-1}(-k) = \pi - x$.