With reference to the definitions of sinhx and coshx show that: cosh^2 x-sinh^2 x =1

Question

sinh x = (e^x-e^-x/2) and cosh x = (e^x + e^-x/2) 

Answer 1

Careful with parenthesis!

sinh(x) = (e^x - e^-x)/2

and

cosh(x) = (e^x + e^-x)/2.

 

We calculate the squares of sinh and cosh, starting with cosh.

cosh²(x) = [(e^x + e^-x)/2]²

             =(e^2x +2 • e^x • e^-x +e^-2x)/4 = (e^2x + 2 + e^-2x)/4

 

sinh²(x) = [(e^x - e^-x)/2]²
             =(e^2x -2 • e^x • e^-x +e^-2x)/4 =(e^2x -2 +e^-2x)/4

Now, we subtract:

cosh²(x) - sinh²(x) =(e^2x +2  +e^-2x)/4

                             -(e^2x -2 +e^-2x)/4

We factor out (/4), because both expressions for cosh and sinh are divided by it,

and open the second parenthesis, which effectively will change the sign of all terms in these parenthesis.

I organized in this fashion, so you see that we subtract all the exponents. Then,

cosh²(x) - sinh²(x) = (e^2x +2 +e^-2x
                                -e^2x +2 -e^-2x)/4

                            

                              =(2 + 2)/4 = 4/4 =1.