An electric dipole is placed at the origin along the x-axis. The electric field at any point, whose position vector makes an angle $\theta$ with the x-axis, makes an angle $\beta$with x axis where tan$\alpha$=$\frac{1}{2}\tan \theta$. Then $\beta$is 

If resultant electric field is making an angle $\alpha$ with the position vector then

$\tan \alpha =\frac{\frac{1}{4\pi {\epsilon }_0}\frac{P\sin \theta }{{\mathrm{r}}^3}}{\frac{1}{4\pi {\epsilon }_0}\frac{2\mathrm{Psin}\theta }{{\mathrm{r}}^3}}=\frac{\tan \theta }{2}$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{net} }\mathrm{makes} \mathrm{an} \mathrm{angle} \mathrm{\alpha } \mathrm{with} \mathrm{the} \mathrm{position} \mathrm{vector} \mathrm{of} \mathrm{the} \mathrm{point}, \\ \tan \left(\mathrm{\alpha }\right)=\tan \left(\frac{\mathrm{\theta }}{2}\right) \\ \mathrm{where} \mathrm{\theta } \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{made} \mathrm{by} \mathrm{the} \mathrm{position} \mathrm{vector} \mathrm{of} \mathrm{the} \\ \mathrm{point} \mathrm{with} \mathrm{axis} \mathrm{of} \mathrm{dipole}. \\ \mathrm{So}, \mathrm{the} \mathrm{angle} \mathrm{made} \mathrm{by} {\mathrm{E}}_{\mathrm{net}} \mathrm{with} \mathrm{x}-\mathrm{axis} \beta \mathrm{is} \mathrm{\alpha }+\mathrm{\theta } \end{gathered}$$