Initial 36. $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ of $\triangle A B C$ respectively, and $\frac{\operatorname{ctg} B+\operatorname{ctg} C}{\operatorname{ctg} A}=3990$. Find the value of $\frac{b^{2}+c^{2}}{a^{2}}$.
Solving, by the cosine theorem we get
$$
\begin{array}{l}
a^{2}+b^{2}-c^{2}=2 a b \cos C=2 a b \sin C \cdot \operatorname{ctg} C \\
=4 S \operatorname{ctg} C .(S \text { is the area of the triangle) }
\end{array}
$$

Similarly, we get $c^{2}+a^{2}-b^{2}=4 S \operatorname{ctg} B$,
$$
b^{2}+c^{2}-a^{2}=4 S \operatorname{ctg} A \text {. }
$$
(1) + (2) gives $\operatorname{ctg} B+\operatorname{ctg} C=\frac{a^{2}}{2 S}$.

From (3) and (4), we get $\frac{\operatorname{ctg} B+\operatorname{ctg} C}{\operatorname{ctg} A}=\frac{2 a^{2}}{b^{2}+c^{2}-a^{2}}$.
Thus, $\frac{2 a^{2}}{b^{2}+c^{2}-a^{2}}=3990$. Simplifying, we get $\frac{b^{2}+c^{2}}{a^{2}}=\frac{1996}{1995}$.