Question

proof arcsech x= arccosh 1/x

Answer 1

here's my work
\[y=arcsech x\]\[x=sech y\]\[x=1/\cosh y\],
im not sure if i need to use comparison here.

Answer 2

are there any restrictions on y? I don't remember for the hyperbolic trig functions but I know there are restriction on the trig functions.

e.g. if y = arctan x => -pi/2 < y < pi/2

Answer 3

Your steps look great so far!
You need to take it a tiny bit further,
multiply both sides by cosh y, divide by x, ya?

Answer 4

@zepdrix i got coshy=1/x

Answer 5

Good :) inverse again, ya?

Answer 6

i see, thanks

Answer 7

And then the only minorly confusing thing to remember is that
this y you end up with is the same y that you started with,

so that's how you connect the arcsech x to the arccosh1/x

Answer 8

noted, thank again