$\newcommand\th x$As in my previous answers, use the substitution 
$$t=\tan\frac\th4,\quad \sin\frac\th2=\frac{2t}{1+t^2},
\quad \cos\frac\th2=\frac{1-t^2}{1+t^2},
\quad \sin\th=\frac{4t(1-t^2)}{(1+t^2)^2},
\quad \th=4\tan^{-1}t$$
in this case to rewrite the inequality in question as 
\begin{equation*}
	f(t):= \frac{4 t \left(41 t^6-11 t^4-285
   t^2-225\right)}{\left(t^2+1\right)^2 \left(41 t^4-90
   t^2+225\right)}+\tan ^{-1}(t)>0	\tag{10}\label{10}
\end{equation*} 
for all 
\begin{equation*}
	t\in(0,t_*],\quad t_*:=\tan\frac\pi8=\sqrt{2}-1. 
\end{equation*}
One has 
\begin{equation*}
	f'(t)= \frac{-5043 t^{12}+6806 t^{10}+278307 t^8+262740 t^6+11475
   t^4-182250 t^2-151875}{\left(t^2+1\right)^3 \left(41 t^4-90
   t^2+225\right)^2}<0
\end{equation*} 
for $t\in(0,t_*)$. 

So, in fact we have the inequality opposite to \eqref{10}: 
\begin{equation*}
	f(t)<0
\end{equation*} 
for $t\in(0,t_*]$.