$\newcommand\th x$As in my previous answers [1][1] and [2][2] to your questions, 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{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{32 t^6 \left(41 t^4-2490
   t^2-2175\right)}{\left(t^2+1\right)^3 \left(41 t^4-90
   t^2+225\right)^2}>0
\end{equation*} 
for $t\in(0,t_*)$. 

Also, $f(0+)=0$. So, \eqref{10} immediately follows. 

---

Note that 
\begin{equation*}
	f(t)= \frac{3712t^7}{4725}+O(t^9). 
\end{equation*}
So, the (lower) rational approximation of $\tan ^{-1}(t)$ given by \eqref{10}) may seem impressive; however, it it far from the best of its kind. Indeed, using Padé approximation, we get 
\begin{equation*}
	g(t)= \frac{16384 t^{17}}{703956825}+O(t^{19}), 
\end{equation*}
where 
\begin{equation*}
	g(t):= \tan ^{-1}(t)-\frac{t \left(15159 t^6+147455 t^4+345345
   t^2+225225\right)}{35 \left(35 t^8+1260 t^6+6930 t^4+12012
   t^2+6435\right)}, \tag{20}\label{20}
\end{equation*}
so that we have a much better rational approximation of $\tan ^{-1}(t)$ with the same degrees of the numerator and the denominator as in the rational expression in \eqref{10}. 

Moreover, 
\begin{equation*}
	g'(t)= \frac{16384 t^{16}}{\left(t^2+1\right) \left(35 t^8+1260 t^6+6930
   t^4+12012 t^2+6435\right)^2}>0
\end{equation*} 
for real $t>0$. 
Also, $g(0+)=0$. So, $g(t)>0$ for all real $t>0$. That is, the rational expression in \eqref{20} is a lower rational approximation of $\tan ^{-1}(t)$.


  [1]: https://mathoverflow.net/a/464190/36721
  [2]: https://mathoverflow.net/a/475377/36721