Let x = ln(sec(y) + tan(y). Show that sec(y) = cosh (x).?

Remember that cosh(x) = (1/2) * (e^(x) + e^(-x)) ..1

and as sec^2 y- tan ^2 y = 1 so 1/(sec y + tan y) = sec y - tan y ..2

x = ln(sec(y) + tan(y))

so e^x = sec(y) + tan(y) ...3
so e^-x = 1/( sec y + tan y) = sec y - tan y .. 4 ( from 2)

add (3) and (4) to get e^x + e^-x = 2 sec (y)

or sec y = (e^ x + e^-x)/2 = cosh x (from 1)