# How do you calculate tan(arccos(5/13))?

Feb 6, 2015

The answer is: $\frac{12}{5}$.

If you set $\alpha = \arccos \left(\frac{5}{13}\right) \Rightarrow \cos \alpha = \frac{5}{13}$ than we have to calculate:

$\tan \alpha = \sin \frac{\alpha}{\cos} \alpha$.

$\cos \alpha = \frac{5}{13} \Rightarrow \sin \alpha = \sqrt{1 - {\cos}^{2} \alpha} = \sqrt{1 - \frac{25}{169}} =$

$= \sqrt{\frac{169 - 25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.

So:

$\tan \alpha = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{13} \cdot \frac{13}{5} = \frac{12}{5}$.