# How do you evaluate sec^-1(sec((13pi)/10))?

${\sec}^{- 1}$ or $a r c \sec$ is the opposite funtion of $\sec$.
Just like ${\sqrt{x}}^{2} = x \mathmr{and} 5 \times x \div 5 = x$
$= \frac{13 \pi}{10}$