Question 415076
There is a variant of the Pythagorean identity: {{{(sec x)^2 = (tan x)^2 + 1}}}.  Hence {{{(sec 18^o)^2 = (tan 18^o)^2 + 1}}}, and so {{{(sec 18^o)^2 - (tan 18^o)^2 = 1}}}