# How do you evaluate sec^-1(sec((9pi)/10))?

${\sec}^{-} 1 \left(\sec \left(\frac{9 \pi}{10}\right)\right) = \frac{9 \pi}{10}$
The range of ${\sec}^{-} 1$ is ${\sec}^{-} 1 \in \left[0 , \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2} , \pi\right]$and since $\frac{9 \pi}{10} \in \left[0 , \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2} , \pi\right]$, ${\sec}^{-} 1 \left(\sec \left(\frac{9 \pi}{10}\right)\right) = \frac{9 \pi}{10}$.