# How do you evaluate sec^-1(-2)?

Jul 11, 2017

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

#### Explanation:

From definition of inverse ratio if $\sec x = - 2$, ${\sec}^{- 1} \left(- 2\right) = x$ and range is $\left[0 , \pi\right\}$

We know that secant function is negative in $Q 2$

as $\sec \left(\frac{\pi}{3}\right) = 2$, we have $\sec \left(\pi - \frac{\pi}{3}\right) = \sec \left(\frac{2 \pi}{3}\right) = - 2$

Hence ${\sec}^{- 1} \left(- 2\right) = \frac{2 \pi}{3}$