写出 $\tan^{-1} p$ 的值域。 已知 $\cos^{-1} k = x$，其中 $0 \le x \le \frac{\pi}{2}$，求 $\cos^{-1}(-k)$ 的主值。

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, $\cos^{-1}(-k) = \pi - x$.