Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 7

Let's recall the definitions: tanh(x) = (sinh(x))/(cosh(x)), se ch(x) = 1/(cosh(x)). Also, cosh(x) = (e^x + e^(-x))/2 and sinh(x) = (e^x - e^(-x))/2.
Now the left side of our identity may be rewritten as
tanh^2(x) + sec h^2(x) = (sinh^2(x) + 1)/(cosh^2(x)).
While it is well-known that sinh^2(x) + 1 = cosh^2(x), we can prove this directly:
sinh^2(x) + 1 = ((e^x - e^(-x))/2)^2 + 1 = ((e^x)^2 - 2 + (e^(-x))^2 + 4)/4 =
=((e^x)^2 + 2 + (e^(-x))^2)/4 =((e^x + e^(-x))/2)^2 = cosh^2(x).
This way the left side is equal to 1, which is the right side. This way the identity is proved.