Let $$x = \frac{R}{2}\;\log\left(\frac{1 + \sqrt{u^2 + v^2}}{1 - \sqrt{u^2 + v^2}}\right)$$ from which I derived that $$\tanh \frac{x}{R} = \sqrt{u^2 + v^2}$$ I have difficulty somehow in deriving the formula $$\cosh \frac{x}{R} = \frac{1}{\sqrt{1- u^2 - v^2}}$$ Any hints?
Asked
Active
Viewed 65 times
1
-
Knowing $\tanh$, you can find $\operatorname{sech}$, and then $\cosh$. – Blue Apr 17 '21 at 18:18
-
$1-\tanh^2(u)=\frac 1{\cosh^2(u)}$ – zwim Apr 17 '21 at 18:31
-
Right, thank you very much – cip Apr 17 '21 at 18:40