Hyperbolic square difference identity

\cosh^2x - \sinh^2x = 1

We know that:
- \sinh x = \frac{e^x - e^{-x}}{2}
- \cosh x = \frac{e^x + e^{-x}}{2}
Square both functions:
- \sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}
- \cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}
\cosh^2 x - \sinh^2 x
- = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}
- = \frac{e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}}4
- = \frac44
- =1
So we’ve shown that \cosh^2x - \sinh^2x = 1!

Question	Answer
What is the identity relating \cosh^2x and \sinh^2x?	\cosh^2x - \sinh^2x = 1
How is \sinh x expressed in exponential form?	\sinh x = \frac{e^x - e^{-x}}{2}
How is \cosh x expressed in exponential form?	\cosh x = \frac{e^x + e^{-x}}{2}
What is the expanded form of \sinh^2 x?	\sinh^2 x = \frac{e^{2x} - 2 + e^{-2x}}{4}
What is the expanded form of \cosh^2 x?	\cosh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4}
Show the proof of \cosh^2x - \sinh^2x = 1.	\cosh^2 x - \sinh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac44 = 1