I Is arcsec = 1 divided by arccos ? Arcsec = 1/arccos

[PeroK]

Its completely different. Not sure what I am doing wrong. Unless arccos(x) ≠arcsec(1/x) like I stated at the top right of the board.

If you insist on your own idiosynchratic approach to trig identities, then I think the onus is on you to sort out the problems. I would encourage you to study this derivation. Let $x = \cos y$, with $y \in [0, \pi]$ :
$$\arccos(x) = \arccos(\cos y) = y$$ $$arcsec(\frac 1 x) = arcsec(\sec y) = y$$ QED

 

[Vividly]

Its completely different. Not sure what I am doing wrong. Unless arccos(x) ≠arcsec(1/x) like I stated at the top right of the board.

I think I know what I did wrong. I redid the problem and got the right answer. I just don’t understand how when you put values in the calculator for arccosx and arcsec(1/x) you get and error for one and a number for the other.

 

[PeroK]

I think I know what I did wrong. I redid the problem and got the right answer. I just don’t understand how when you put values in the calculator for arccosx and arcsec(1/x) you get and error for one and a number for the other.

The calculator on my phone doesn't have an option for $arcsec$; nor does excel (it seems); nor does $arcsec$ exist in Latex. None of that matters in light of post #31.

 

[Vividly]

The calculator on my phone doesn't have an option for $arcsec$; nor does excel (it seems); nor does $arcsec$ exist in Latex. None of that matters in light of post #31.

This youtube video teaches you how to put it in the calculator.

 

[PeroK]

I like that equation:$$\sec^{-1}(-2) = \cos^{-1}(-\frac 1 2) = \frac {2\pi}{3}$$