Is this true (Natural Logarithm)?

Question

Is it true if I say, when we know that $\text{D},\beta\in\mathbb{R}$:

$$\text{D}=e^{-\pi\frac{\beta}{\sqrt{1-\beta^2}}}\Longleftrightarrow \beta=-\frac{\ln(\text{D})}{\sqrt{\pi^2+\ln^2(\text{D})}},0\le\text{D}<1$$

Answer 1

For $D,\beta\in\mathbb R$ such that $0\le D\lt 1$, we can say that $$D=e^{-\pi\frac{\beta}{\sqrt{1-\beta^2}}}\iff \beta=-\frac{\ln(D)}{\sqrt{\pi^2+\ln^2(D)}} .$$

Proof for $\Rightarrow$ :

$$\begin{align}&0\le D\lt 1\quad\text{and}\quad D=e^{-\pi\frac{\beta}{\sqrt{1-\beta^2}}} \\&\Rightarrow 1-\beta^2\gt 0\quad\text{and}\quad 0\le e^{-\pi\frac{\beta}{\sqrt{1-\beta^2}}}\lt 1\quad\text{and}\quad \ln(D)=-\pi\frac{\beta}{\sqrt{1-\beta^2}} \\&\Rightarrow -1\lt \beta\lt 1\quad\text{and}\quad -\pi\frac{\beta}{\sqrt{1-\beta^2}}\lt 0\quad\text{and}\quad \ln^2(D)=\frac{\pi^2\beta^2}{1-\beta^2} \\&\Rightarrow -1\lt\beta\lt 1\quad\text{and}\quad \beta\gt 0\quad\text{and}\quad \beta^2=\frac{\ln^2(D)}{\pi^2+\ln^2(D)}\\&\Rightarrow \beta=-\frac{\ln(D)}{\sqrt{\pi^2+\ln^2(D)}}\end{align}$$

Proof for $\Leftarrow$ :

$$\begin{align}&0\le D\lt 1\quad\text{and}\quad \beta=-\frac{\ln(D)}{\sqrt{\pi^2+\ln^2(D)}} \\&\Rightarrow 0\le D\lt 1\quad\text{and}\quad \beta=-\frac{\ln(D)}{\sqrt{\pi^2+\ln^2(D)}}\quad\text{and}\quad \beta\gt 0 \\&\Rightarrow 0\le D\lt 1\quad\text{and}\quad \beta^2=\frac{\ln^2(D)}{\pi^2+\ln^2(D)}\quad\text{and}\quad \beta\gt 0 \\&\Rightarrow 0\le D\lt 1\quad\text{and}\quad \ln^2(D)=\frac{\beta^2\pi^2}{1-\beta^2}\quad\text{and}\quad 1-\beta^2\gt 0\quad\text{and}\quad \beta\gt 0 \\&\Rightarrow 0\le D\lt 1\quad\text{and}\quad \ln(D)=-\frac{\beta\pi}{\sqrt{1-\beta^2}}\quad\text{and}\quad 0\lt\beta\lt 1 \\&\Rightarrow D=e^{-\pi\frac{\beta}{\sqrt{1-\beta^2}}} \end{align}$$