Let $f(x) = \log(\cosh(x))$, and define the kernel density:
$$p_r(\phi;\theta) = \Big(f\big(r\cos(\phi-\theta)\big) - f\big(r\cos(\phi+\theta)\big) \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0\le \phi,\theta \le \mbox{$\large\frac{\pi}{2}$},\;\,r>0$$
Show that for all $r>0$, $p_r(\phi;\theta)$ has a decreasing likelihood ratio. I.e., for $0\le\theta_1 < \theta_2\le\pi/2$, the function:
$$ h(\phi) = \frac{f\big(r\cos(\phi-\theta_2)\big) - f\big(r\cos(\phi+\theta_2)\big) }{f\big(r\cos(\phi-\theta_1)\big) - f\big(r\cos(\phi+\theta_1)\big)}$$
is decreasing on $[0,\pi/2]$, and:
$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \log p_r(\phi;\theta) \le 0,\quad 0 \le \phi,\theta\le \mbox{$\large\frac{\pi}{2}$}$$
From the derivative expression, after changing variables and simplifying, I believe the result will be true if for $y>x\ge 0$ :
$$\bigg(\frac{y^2-x^2}{\cosh^2(x)}-x\tanh(x)-y\tanh(y)\!\bigg)\log\!\bigg(\frac{\cosh(y)}{\cosh(x)}\bigg) + \tanh^2(x)\big(y^2-x^2\big) > 0 $$
Edit with progress: Using the bound:
$$\log(\cosh(y))-\log(\cosh(x)) \ge \sqrt{y^2+1}-\sqrt{x^2+1}$$
with just the positive $(y^2-x^2)/\cosh^2(x)$ term, we can isolate the $\log \cosh$ term, producing the following inequality. For $y>x\ge 0$ :
$$\frac{\big(\sqrt{y^2+1}-\sqrt{x^2+1}+\sinh^2(x)\big)\big(y^2-x^2\big)}{\cosh^2(x)\big(x\tanh(x)+y\tanh(y)\big)} - \log\!\bigg(\frac{\cosh(y)}{\cosh(x)}\bigg) \;>\; 0$$
This function is increasing in $y$, taking the derivative with respect to $y$ would eliminate $\log \cosh$, and it would suffice to show that this expression is positive. However the derivative expression does not seem to simplify easily.
This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.
