arsinh ⁡ ( tan ⁡ α ) = artanh ⁡ ( sin ⁡ α ) = ln ⁡ ( 1 + sin ⁡ α cos ⁡ α ) = ± arcosh ⁡ ( 1 cos ⁡ α ) {\displaystyle \operatorname {arsinh} \left(\tan \alpha \right)=\operatorname {artanh} \left(\sin \alpha \right)=\ln \left({\frac {1+\sin \alpha }{\cos \alpha }}\right)=\pm \operatorname {arcosh} \left({\frac {1}{\cos \alpha }}\right)} ln ⁡ ( | tan ⁡ α | ) = − artanh ⁡ ( cos ⁡ 2 α ) {\displaystyle \ln \left(\left|\tan \alpha \right|\right)=-\operatorname {artanh} \left(\cos 2\alpha \right)} [9]