已知 $\sin A = p$，其中 $A$ 为钝角，求：

1. Use the Pythagorean identity to express $\cos^2 A$ in terms of $p$

$\sin^2 A + \cos^2 A = 1$
$\cos^2 A = 1 - \sin^2 A$
$\cos^2 A = 1 - p^2$

2. Determine the sign of $\cos A$ using the given quadrant

Since $A$ is obtuse ($90^\circ < A < 180^\circ$), $\cos A$ must be negative.
$\cos A = -\sqrt{1 - p^2}$

1. Express $\cot A$ in terms of $\sin A$ and $\cos A$

$\cot A = \frac{\cos A}{\sin A}$

2. Substitute the values of $\sin A$ and $\cos A$ to find $\cot A$

From part (i), $\cos A = -\sqrt{1 - p^2}$.
$\cot A = \frac{-\sqrt{1 - p^2}}{p}$