If ψ(ϕ)=ln(secϕ+tanϕ), how do you find an expression for ϕ?

Question

It should be $2\tan ^{-1}\left( e^{\psi }\right) -\dfrac{\pi }{2}$ but i'm not sure whether that is correct or if the $\dfrac{\pi }{2}$ should be in brackets. It would be helpful if you also found an expression for $\phi$ in terms of hyperbolic trig functions

Answer 1

$$\psi=\ln(\sec\phi+\tan\phi)\rightarrow e^\psi=\sec\phi+\tan\phi$$ Now, using $\tan^2\psi+1=\frac{1}{\cos^2\phi}$ and working in $[0,\pi/2]$ $$e^\psi=\frac{1}{\cos\phi}+\sqrt{\frac{1}{\cos^2\phi}-1}=\frac{1}{\cos\phi}\left(1+\sqrt{1-\cos^2\phi}\right)=\frac{1}{\cos\phi}\left(1+\sin\phi\right)$$ $$e^\psi\cos\phi-1=\sin\phi$$ $$e^{2\psi}\cos^2\phi+1-2e^\psi\cos\phi=\sin^2\phi=1-\cos^2\phi$$ $$(e^{2\psi}+1)\cos^2\phi-2e^\psi\cos\phi=0$$ $$\cos\phi((e^{2\psi}+1)\cos\phi-2e^\psi)=0$$ So $$\cos\phi=0 $$ that is excluded by the existence condition or $$\cos\phi=\frac{2e^\psi}{1+e^{2\psi}}$$ $$\psi=\cos^{-1}\frac{2e^\psi}{1+e^{2\psi}}=\cos^{-1}\frac{2}{e^{-\psi}+e^{\psi}}=\cos^{-1}\left({\frac{1}{\cosh{\psi}}}\right)$$