Does this hyperbolic identity actually exist?

Question

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

I seen it in a textbook and can not seem to prove it.

Answer 1

The identity $\sinh^2(x) + \cosh^2(x) + 1 = 2\cosh(x)$ is not correct.

We have

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

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

FRED