10. (14 points) Given a real number $a>1$. Find the minimum value of the function
$$
f(x)=\frac{(a+\sin x)(4+\sin x)}{1+\sin x}
$$
10. Note that,
$$
\begin{array}{l}
f(x)=\frac{(a+\sin x)(4+\sin x)}{1+\sin x} \\
=1+\sin x+\frac{3(a-1)}{1+\sin x}+a+2 .
\end{array}
$$

When $0 < t \leq 2$, the function
$$
y=t+\frac{3(a-1)}{t}
$$

is decreasing in $(0,2]$, thus,
$$
\begin{array}{l}
f(x)_{\text {min }}=f(1)=2+\frac{3(a-1)}{2}+a+2 \\
=\frac{5(a+1)}{2} .
\end{array}
$$