Find in terms of $p$, $\tan(-\alpha)$, $\tan(\pi - \alpha)$ and $\tan(\frac{\pi}{2}-\alpha)$. Given that $\tan$ $\alpha = p$, where $\alpha$ is acute, find in terms of $p$, $\tan(-\alpha)$, $\tan(\pi - \alpha)$ and $\tan(\frac{\pi}{2}-\alpha)$.
Hint: Use the following facts:


$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$ 
$\sin(-\alpha) = -\sin(\alpha)$
$\cos(-\alpha) = \cos(\alpha)$
$\sin(\pi-\alpha) = \sin(\alpha)$
$\cos(\pi-\alpha) = -\cos(\alpha)$
$\sin(\frac{\pi}{2}-\alpha) = \cos(\alpha)$
$\cos(\frac{\pi}{2}-\alpha) = \sin(\alpha)$.
$$\cos\dfrac{\pi}{2+\alpha} = - \sin \alpha \\
\sin\dfrac{\pi}{2+\alpha} = \cos \alpha$$