Example 4. (Jiehu City 1984 Hua Math Competition Question) Let the three sides of $\triangle ABC$ be $a, b, c$, and $\frac{1}{c-a}-\frac{1}{c+b}=1$. What kind of triangle is $\triangle ABC$?
From the given, we can derive that $b(c+b)-a(c-a)$ $=(c-a)(c+b)$,
which means $b^{2}+a^{2}-c^{2}=-a b$, and $\frac{b^{2}+a^{2}-c^{2}}{2 a b}=-\frac{1}{2}$. Therefore, $\cos C=-\frac{1}{2}$. Hence, this triangle is an obtuse triangle.