# How do you calculate tan(arccos(-9/10))?

May 1, 2015

In this way.

We have to calculate $\tan \alpha$, where alpha is:

$\alpha = \arccos \left(- \frac{9}{10}\right)$.

Since the range of the function $y = \arccos x$ is $\left[0 , \pi\right]$ and the value negative $- \frac{9}{10}$, the angle is in the second quadrant, in which the sinus is positive. So:

$\cos \alpha = - \frac{9}{10}$,

$\sin \alpha = + \sqrt{1 - {\cos}^{2} \alpha} = + \sqrt{1 - \frac{81}{100}} = \frac{\sqrt{19}}{10}$.

Than:

$\tan \alpha = \sin \frac{\alpha}{\cos} \alpha = \frac{\frac{\sqrt{19}}{10}}{- \frac{9}{10}} = - \frac{\sqrt{19}}{10} \cdot \frac{10}{9} =$

$= - \frac{\sqrt{19}}{9}$.