How do you prove (secx-tanx)(secx+tanx) =secx ?

The given identity is false.
##(sec(x) – tan(x))(sec(x) + tan(x)) = 1##
We will be using the following:
##sec(x) = 1/cos(x)## (by definition)
##tan(x) = sin(x)/cos(x)## (by definition)
##(a-b)(a+b) = a^2 – b^2## (difference of squares formula)
##sin^2(x) + cos^2(x) = 1## (identity)
##(sec(x) – tan(x))(sec(x) + tan(x)) = sec^2(x) – tan^2(x)##
(by the difference of squares formula)
##= (1/cos(x))^2 – (sin(x)/cos(x))^2##
(by definition of secant and tangent)
##= 1/cos^2(x) – sin^2(x)/cos^2(x)##
##=(1 – sin^2(x))/cos^2(x)##
##= cos^2(x)/cos^2(x)##
(by the above identity)
##=1##