Is it true that $\frac{1}{\cosh(x) - \sinh(x)} = e^{x}$?

Question

Is it true that:

$$\frac{1}{\cosh(x) - \sinh(x)} = e^{x}$$

Graphically, it seems to be true, but I am not sure if it is true for all $x$.

Also, if it is, is it a known result?

Answer 1

$$\cosh(x)=\frac{e^x+e^{-x}}{2}$$

$$\sinh(x)=\frac{e^x-e^{-x}}{2}$$

$$\cosh(x)-\sinh(x)=e^{-x}$$

$$\frac{1}{\cosh(x)-\sinh(x)}=e^{x}$$

Answer 2

Recall:

• $e^x = \cosh x + \sinh x$.
• $(\cosh x)^2 - (\sinh x)^2 =1$.

And do not forget $a^2 - b^2 = (a-b)(a+b)$.

Answer 3

Just use $\cosh{x}=\tfrac12(e^x+e^{-x})$ and $\sinh{x}=\tfrac12(e^{x}-e^{-x})$

Answer 4

Yep. Use the well-known identities: $$ \cosh x = \frac{e^x + e^{-x}}{2} \textrm{ and } \sinh x = \frac{e^x - e^{-x}}{2} $$

This is sort of analogous to $\cos x - i \sin x = e^{-ix}$.

Answer 5

$$ \mathrm{cosh} x = \dfrac{e^{x}+e^{-x}}{2}, \ \mathrm{sinh} x = \dfrac{e^{x}-e^{-x}}{2} $$ It follows from this and the fact that the denominator is non-zero for all real $x$, that the equality is true.

Answer 6

We have, $\cosh(x)=\frac{e^x+e^{-x}}{2}$, also, $\sinh(x)=\frac{e^x-e^{-x}}{2}$. So, $\cosh(x)-\sinh(x)=e^{-x}$ or $\frac{1}{\cosh(x)-\sinh(x)}=e^x$. Yes, it is well known.