# How do you simplify  tan[cos^-1(-(sqrt3/2))]?

Jul 22, 2018

$- \frac{1}{\sqrt{3}}$.

#### Explanation:

The cosine value in brackets being negative,

${\cos}^{- 1} \left(- \frac{\sqrt{3}}{2}\right) \in \left[\frac{\pi}{2} , \pi\right]$. So,

$\tan {\cos}^{- 1} \left(- \frac{\sqrt{3}}{2}\right)$

$= \tan \left({\cos}^{- 1} \cos \left(\pi - \frac{\pi}{6}\right)\right)$

$= \tan \left(\pi - \frac{\pi}{6}\right)$

$= - \tan \left(\frac{\pi}{6}\right)$

$= - \frac{1}{\sqrt{3}}$

Jul 22, 2018

-0.5774

#### Explanation:

There is an Inverse Trigonometric Identity that will help on this webpage
https://brilliant.org/wiki/inverse-trigonometric-identities/

It says that ${\cos}^{- 1} \left(- x\right) = \pi - {\cos}^{- 1} \left(x\right)$

Using that, your expression can be simplified to

$\tan \left(\pi - {\cos}^{-} 1 \left(\frac{\sqrt{3}}{2}\right)\right)$

Do you need to take it further? Taking just part of the above

${\cos}^{-} 1 \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$

Plugging that into $\tan \left(\pi - {\cos}^{-} 1 \left(\frac{\sqrt{3}}{2}\right)\right)$ we get

$\tan \left(\pi - \frac{\pi}{6}\right) = \tan \left(\frac{5 \pi}{6}\right)$

Using my calculator, the whole thing is equal to -0.5774...

I hope this helps,
Steve