## Trigonometry (10th Edition)

$tan (-900^{\circ}) = \frac{sin (-900^{\circ})}{cos (-900^{\circ})} = \frac{0}{-1} = 0$
$sin (\theta) = \frac{y}{r}$ $cos (\theta) = \frac{x}{r}$ We can express $tan(\theta)$ using $sin(\theta)$ and $cos(\theta)$: $tan (\theta) = \frac{y}{x}$ $tan (\theta) = \frac{y/r}{x/r}$ $tan (\theta) = \frac{sin (\theta)}{cos (\theta)}$ Therefore, if we know $sin (\theta)$ and $cos (\theta)$, then we can find $tan(\theta)$. $tan (-900^{\circ}) = \frac{sin (-900^{\circ})}{cos (-900^{\circ})} = \frac{0}{-1} = 0$