Solve 1+p^2/2p=secy

$$\frac { 1 + p ^ { 2 } } { 2 p } = \sec y$$

Solve for p (complex solution)

$$p=\frac{\sqrt{\left(\sin(y)\right)^{2}}+1}{\cos(y)}$$

$$p=\frac{-\sqrt{\left(\sin(y)\right)^{2}}+1}{\cos(y)}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}$$

Solve for p

$$p=\frac{\cos(y)\sqrt{-\left(\cos(y)\right)^{2}+1}+|\cos(y)|}{\cos(y)|\cos(y)|}$$

$$p=\frac{-\cos(y)\sqrt{-\left(\cos(y)\right)^{2}+1}+|\cos(y)|}{\cos(y)|\cos(y)|}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}$$