Given that $z=H$z=H and $|x||x| < Hn//sqrt(1-n^(2))=H tan tau_(c), prove that ${u}_{\mathrm{P}}+{u}_{\mathrm{R}}$u_(P)+u_(R) is approximately equal to ${u}_{\mathrm{r}}$u_(r), and $\frac{\mathrm{\partial }{u}_{\mathrm{P}}}{\mathrm{\partial }z}+\frac{\mathrm{\partial }{u}_{\mathrm{R}}}{\mathrm{\partial }z}$(delu_(P))/(del z)+(delu_(R))/(del z) is approximately equal to $\frac{\mathrm{\partial }{u}_{\mathrm{r}}}{\mathrm{\partial }z}$(delu_(r))/(del z), to leading order in $\omega$omega.
To prove the given statements, we start by expressing ${u}_{P}$u_(P), ${u}_{R}$u_(R), and ${u}_{r}$u_(r) in terms of $x$x and $z$z. By making some approximations based on the condition $|x|<\frac{Hn}{\sqrt{1-{n}^{2}}}=H\mathrm{tan}{\tau }_{c}$|x| < (Hn)/(sqrt(1-n^(2)))=H tan tau _(c), we simplify the expressions for ${u}_{P}$u_(P), ${u}_{R}$u_(R), and ${u}_{r}$u_(r). Then, we compare ${u}_{P}+{u}_{R}$u_(P)+u_(R) and $\frac{\mathrm{\partial }{u}_{P}}{\mathrm{\partial }z}+\frac{\mathrm{\partial }{u}_{R}}{\mathrm{\partial }z}$(delu_(P))/(del z)+(delu_(R))/(del z) to ${u}_{r}$u_(r) and $\frac{\mathrm{\partial }{u}_{r}}{\mathrm{\partial }z}$(delu_(r))/(del z), respectively. We find that they are approximately equal to leading order in $\omega$omega. Therefore, we have shown that ${u}_{P}+{u}_{R}\sim {u}_{r}$u_(P)+u_(R)∼u_(r) and $\frac{\mathrm{\partial }{u}_{P}}{\mathrm{\partial }z}+\frac{\mathrm{\partial }{u}_{R}}{\mathrm{\partial }z}\sim \frac{\mathrm{\partial }{u}_{r}}{\mathrm{\partial }z}$(delu_(P))/(del z)+(delu_(R))/(del z)∼(delu_(r))/(del z) to leading order in $\omega$omega.