Help with algebra equation!?

Question

Given that f(x)=(e^x+e^-x) / 2 and g(x)=(e^x-e^-x) / 2 ,show that [f(x)]^2-[g(x)]^2=1

Answer 1

Let f(x) = (e^x+e^-x) / 2

      g(x) = (e^x+e^-x) / 2

To prove:

                We have to prove that [f(x)]^2 - [g(x)]^2 = 1

Proof:

          [f(x)]^2 = {[e^x + e^-x]/2}^2

                      = {[e^x + e^-x]^2 }/ 2^2

                      = [e^2x + e^-2x +2e^x.e^-x] / 4        [since (a+b)^2 = a^2 + b^2 + 2ab]

                      = [e^2x + e^-2x +2e^(x-x)] / 4           [since a^m.a^n = a^(m+n)]

                      = [e^2x + e^-2x + 2e^0] / 4

                      = [e^2x + e^-2x + 2.1] / 4                  [since e^0 = 1]

                      = [e^2x + e^-2x + 2] / 4    

      [g(x)]^2 = {[e^x - e^-x]/2}^2

                      = {[e^x - e^-x]^2 }/ 2^2

                      = [e^2x + e^-2x -2e^x.e^-x] / 4        [since (a-b)^2 = a^2 + b^2 - 2ab]

                      = [e^2x + e^-2x -2e^(x-x)] / 4           [since a^m.a^n = a^(m+n)]

                      = [e^2x + e^-2x -2e^0] / 4

                      = [e^2x + e^-2x -2.1] / 4                  [since e^0 = 1]

                      = [e^2x + e^-2x - 2] / 4  

[f(x)^2]^2 - [g(x)^2]^2 = { [e^2x + e^-2x + 2] / 4}  - { [e^2x + e^-2x - 2] / 4 }

                                    =  (1/4) [e^2x + e^-2x +2] - (1/4) [e^2x + e^-2x -2]

                                    =  (1/4) [e^2x + e^-2x +2 -e^2x - e^-2x +2]

                                    =  (1/4) [ 0+4]

                                    =  4/4

                                    = 1

Hence it is proved that [f(x)]^2 - [g(x)]^2 = 1.