The relation between hyperbolic sine and hyperbolic cotangent

Question

I was wondering if someone can verify (or not) the correctness of the following function?

$$\frac{1}{\sinh^2X}=\coth^2X-1$$

I saw it in a paper but I am weak in math, so I am unsure if it is correct or not. I assume that the author took the following two functions in consideration:

$$\operatorname{csch}^2X=\coth^2X-1$$ $$\operatorname{csch}X=\frac{1}{\sinh X}$$

Thank you in advance.

Answer 1

$$\frac{1}{\sinh^2(x)}=\underbrace{\frac{\cosh^2(x)}{\sinh^2(x)}}_{=\text{coth}^2(x)}\cdot \underbrace{\frac{1}{\cosh^2(x)}}_{=1-\tanh^2(x)}=\text{coth}^2(x)\left(1-\tanh^2(x)\right)$$$$=\text{coth}^2(x)-\underbrace{\text{coth}^2(x)\cdot \tanh^2(x)}_{=\text{coth}^2(x)\cdot \frac{1}{\text{coth}^2(x)}=1}=\text{coth}^2(x)-1$$

Answer 2

You have the identity $$\text{cosh}^2 x - \text{sinh}^2 x = 1 $$

Multiply this by $\frac{1}{\text{sinh}^2 x}$ and see what happens. See more here.

Answer 3

By definition $$\coth X=\frac{\cosh X}{\sinh X}.$$ So the equality follows immediately from the classic identity $\cosh^2 X - \sinh^2 X = 1$.