10. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the function
$$
y=\frac{225}{4 \sin ^{2} x}+\frac{2}{\cos x}
$$

has a minimum value of
10.68
Since $x \in\left(0, \frac{\pi}{2}\right)$, we have $\sin x>0, \cos x>0$. Let $k>0$, then
$$
\begin{array}{l}
y=\frac{225}{4 \sin ^{2} x}+k \sin ^{2} x+\frac{1}{\cos x}+\frac{1}{\cos x}+ \\
\quad k \cos ^{2} x-k \\
\geqslant 15 \sqrt{k}+3 \sqrt[3]{k}-k .
\end{array}
$$

Equality in (1) holds if and only if
$$
\left\{\begin{array} { l } 
{ \frac { 2 2 5 } { 4 \operatorname { s i n } ^ { 2 } x } = k \operatorname { s i n } ^ { 2 } x , } \\
{ \frac { 1 } { \operatorname { c o s } x } = k \operatorname { c o s } ^ { 2 } x }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin ^{2} x=\frac{15}{2 \sqrt{k}}, \\
\cos ^{2} x=\frac{1}{\sqrt[3]{k^{2}}}
\end{array}\right.\right.
$$

At this point, $\frac{15}{2 \sqrt{k}}+\frac{1}{\sqrt[3]{k^{2}}}=1$.
Let $\frac{1}{k}=t^{6}$, then $2 t^{4}+15 t^{3}-2=0$. And
$$
\begin{array}{l}
2 t^{4}+15 t^{3}-2=2 t^{4}-t^{3}+16 t^{3}-2 \\
=t^{3}(2 t-1)+2(2 t-1)\left(4 t^{2}+2 t+1\right) \\
=(2 t-1)\left(t^{3}+8 t^{2}+4 t+2\right),
\end{array}
$$

Thus, $(2 t-1)\left(t^{3}+8 t^{2}+4 t+2\right)=0$.
Noting that $0<t^{6}=\frac{1}{k} \leqslant 1$, it is easy to see that the only root satisfying the condition is $t=\frac{1}{2}$.

When $t=\frac{1}{2}$, $k=\frac{1}{t^{6}}=64$, and equality in (1) is achieved.
Therefore, the minimum value of the function $y=\frac{225}{4 \sin ^{2} x}+\frac{2}{\cos x}$ is $15 \sqrt{64}+3 \sqrt[3]{64}-64=68$.