\(f(x)=\cfrac{1}{\pi(1+x^2)}\)
for \(-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}\)
then \(\sqrt{|tan(\theta)|}\) has a distribution
\(f(x)=\cfrac{1}{\pi(1+x^4)}|2x|\)
for \(-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}\) in blue below.
These set of peaks for the probability density of \(\sqrt{tan(\theta)}\) are not at \(\cfrac{\pi}{2}\) apart but \(\cfrac{2}{\sqrt[4]{3}}\) apart.
\(\cfrac{2}{\sqrt[4]{3}}=1.5197\,\, \small{rad}=87.071^o\ne90^o\)
From,
\(\cfrac{E_{p\,v}}{\hbar}=\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_v}\)
The probability density of \(1-\sqrt{tan(\theta_v)}\) has been right shifted by \(1\). The peak intensity of \(E_p\) occurs not at \(\theta=0\,\,\small{rad}\) but at \(\theta=1\,\, \small{rad}=57.296^o\).
\(a=\cfrac{c}{x_v}\)
is assumed constant.
Note: Also,
\(\cfrac{E_{p\,v}}{\hbar}=\left\{\sqrt{tan(\theta_v)}-1\right\}\cfrac{c}{x_v}\)
in which case two peaks occurs centered about \(\theta=-1\,\, \small{rad}\), \(\theta=1.5197\,\, \small{rad}\) apart. A notch occurs at \(\theta=-1\,\, \small{rad}\).
