## Precalculus (6th Edition)

$1$
$\frac{\pi}{4}=60^o$ is a special angle that intersects the unit circle at the point $(\frac{\sqrt2}{2}, \frac{\sqrt3}{2})$. Since $\tan{\theta}=\frac{y}{x}$, then $\tan{\frac{\pi}{4}} =\dfrac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$