7. Let $D_{1}, D_{2}, \cdots, D_{2 n}$ be $2 n(n \geqslant 1)$ points on the hypotenuse $B C$ of the right triangle $\triangle A B C$. Denote
$$
\angle D_{i-1} A D_{i}=\alpha_{i}(i=1,2, \cdots, 2 n+1) \text {, }
$$

satisfying $D_{i-1} D_{i}=D_{i} D_{i+1}\left(i=1,2, \cdots, 2 n, D_{0}=B\right.$, $D_{2 n+1}=C$ ). Then
$$
\frac{\sin \alpha_{1} \cdot \sin \alpha_{3} \cdots \cdot \sin \alpha_{2 n+1}}{\sin \alpha_{2} \cdot \sin \alpha_{4} \cdots \cdot \sin \alpha_{2 n}}=
$$
$\qquad$
7. $\frac{1}{2 n+1}$.

Since $B D_{1}=D_{1} D_{2}=\cdots=D_{2 n} C$, therefore, the areas of $\triangle B A D_{1}, \triangle D_{1} A D_{2}, \cdots, \triangle D_{2 n} A C$ are equal, each being $\frac{1}{2 n+1}$ of the area of $\triangle A B C$.
$$
\begin{array}{l}
\text { Hence } \prod_{i=0}^{n} S_{\triangle D_{2 i} A D_{2 i+1}} \\
S_{\triangle D_{0} A D_{2 n+1}} \prod_{i=1}^{n} S_{\triangle D_{2 i-1} A D_{2 i}} \\
=\frac{\prod_{i=0}^{n}\left(A D_{2 i} \cdot A D_{2 i+1} \sin \alpha_{2 i+1}\right)}{A D_{0} \cdot A D_{2 n+1} \prod_{i=1}^{n}\left(A D_{2 i-1} \cdot A D_{2 i} \sin \alpha_{2 i}\right)} \\
\Rightarrow \frac{\sin \alpha_{1} \cdot \sin \alpha_{3} \cdots \cdots \cdot \sin \alpha_{2 n+1}}{\sin \alpha_{2} \cdot \sin \alpha_{4} \cdots \cdot \sin \alpha_{2 n}}=\frac{1}{2 n+1} . \\
\end{array}
$$