$\newcommand\th x$As in my previous answers 1 and 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 $\tan ^{-1}(t)$.
