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##