# How do you evaluate sec^-1(sec((19pi)/10))?

##### 1 Answer
Oct 13, 2016

${\sec}^{-} 1 \left(\sec \left(\frac{19 \pi}{10}\right)\right) = {\sec}^{-} 1 \left(\sec \left(\frac{\pi}{10}\right)\right) = \frac{\pi}{10}$

#### Explanation:

${\sec}^{-} 1 \left(\sec \left(\frac{19 \pi}{10}\right)\right)$

Since the restriction for secant inverse is $\left[0 , \pi\right] , y \ne \frac{\pi}{2}$ we can see that $\frac{19 \pi}{10}$ is in quadrant four which is out of the restriction but since secant is positive in quadrant IV we need to find the argument in quadrant I from our restriction. So the reference angle is $\frac{\pi}{10}$ and therefore the argument in quadrant I is also $\frac{\pi}{10}$.

Applying the property ${f}^{-} 1 \left(f \left(x\right)\right) = x$ we have our answer is $\frac{\pi}{10}$