Can someone give a proof for this hyperbolic rule

Question

can someone please give a proof for this with a little explanation

$$\cosh(x)^2-\sinh(x)^2 = 1$$

Answer 1

Hint: Use $$\cosh(x)=\cos(\mathrm{i}x)$$ $$\sinh(x)=-\mathrm{i}\sin(\mathrm{i}x)$$

to turn it into

$$\cos^2(x)+\sin^2(x)=1$$

Also, I suggest you to read this question and its answers: Can hyperbolic functions be defined in terms of trignometric functions?

Answer 2

$$\cosh^2{x}-\sinh^2x=\left(\frac{e^x+e^{-x}}{2}\right)^2-\left(\frac{e^x-e^{-x}}{2}\right)^2=$$ $$=\frac{e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}}{4}=1.$$

Answer 3

One more attempt:

$\cosh(x) =(1/2)(e^x +e^{-x});$

$\sinh(x) = (1/2)(e^x -e^{-x}).$

$\cosh^2(x) - \sinh^2(x)=$

$(\cosh(x)-\sinh(x))$ $×(\cosh(x) +\sinh(x))=$

$(e^{-x})(e^x)=1$.