# Question #da001

Jun 5, 2017

$\sin \theta = \pm \sqrt{1 - {p}^{2}}$
$\tan \theta = \frac{\pm \sqrt{1 - {p}^{2}}}{p}$

#### Explanation:

$\text{given } \cos \theta = p$

$\Rightarrow \sin \theta = \pm \sqrt{1 - {\cos}^{2} \theta}$

$\textcolor{w h i t e}{\Rightarrow \sin \theta} = \pm \sqrt{1 - {p}^{2}}$

$\text{the sign is dependent on which quadrant " theta" is in}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\Rightarrow \tan \theta = \frac{\pm \sqrt{1 - {p}^{2}}}{p}$
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