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

1 Answer
Dec 13, 2015

The given identity is false.

#(sec(x) - tan(x))(sec(x) + tan(x)) = 1#

Explanation:

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#