# How do you evaluate tan^-1(tan((5pi)/6))?

Aug 4, 2016

$= 5 \frac{\pi}{6}$

#### Explanation:

${\tan}^{-} 1 \left(\tan \left(5 \frac{\pi}{6}\right)\right)$
$= 5 \frac{\pi}{6}$

Aug 4, 2016

${\tan}^{- 1} \left(\tan \left(\frac{5 \pi}{6}\right)\right) = - \frac{\pi}{6}$

#### Explanation:

$\theta = {\tan}^{- 1} \left(\tan \left(\frac{5 \pi}{6}\right)\right)$ by definition satisfies both of the conditions:

• $\textcolor{w h i t e}{X} \tan \theta = \tan \left(\frac{5 \pi}{6}\right)$

• $\textcolor{w h i t e}{X} - \frac{\pi}{2} < \theta < \frac{\pi}{2}$

Note that $\tan$ has period $\pi$, so for any integer $n$:

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

When $n = - 1$, we have:

$\frac{5 \pi}{6} + n \pi = \frac{5 \pi}{6} - \pi = - \frac{\pi}{6}$

which lies in the range $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, so satisfies the second condition for ${\tan}^{- 1}$

Thus:

${\tan}^{- 1} \left(\tan \left(\frac{5 \pi}{6}\right)\right) = - \frac{\pi}{6}$