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

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Answer 1 · 2018-07-04T15:59:39

$sec^-1(sec((9pi)/10))=(9pi)/10$

The range of $sec^-1$ is $sec^-1in[0,pi/2)uu(pi/2,pi]$ and since $(9pi)/10in[0,pi/2)uu(pi/2,pi]$ , $sec^-1(sec((9pi)/10))=(9pi)/10$ .

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