User:TeeMee124

{\displaystyle {\begin{aligned}\psi (\phi )&=\ln \left[\tan \left({\frac {\pi }{4}}+{\frac {\phi }{2}}\right)\right]+{\frac {e}{2}}\ln \left[{\frac {1-e\sin \phi }{1+e\sin \phi }}\right]\\&=\tanh ^{-1}(\sin \phi )-e\tanh ^{-1}(e\sin \phi )\\&=\operatorname {gd} ^{-1}(\phi )-e\tanh ^{-1}(e\sin \phi ).\end{aligned}}}