Simplify the following expression as much as possible:

$$
\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^{2}}-1+a}\right)\left(\sqrt{\frac{1}{a^{2}}-1}-\frac{1}{a}\right)
$$

where $0<a<1$.
Let's denote the two terms of the first factor by $A$ and $B$, and the second factor by $C$. Then, by rationalizing the denominators, we have

$$
\begin{aligned}
& A=\frac{\sqrt{1+a}(\sqrt{1+a}+\sqrt{1-a})}{(1+a)-(1-a)}=\frac{1+a+\sqrt{1-a^{2}}}{2 a} \\
& B=\frac{(1-a)\left[\sqrt{1-a^{2}}+(1-a)\right]}{1-a^{2}-(1-a)^{2}}=\frac{(1-a)^{2}+(1-a) \sqrt{1-a^{2}}}{2 a(1-a)}=\frac{1-a+\sqrt{1-a^{2}}}{2 a}
\end{aligned}
$$

and thus

$$
A+B=\frac{2+2 \sqrt{1-a^{2}}}{2 a}=\frac{\sqrt{1-a^{2}}+1}{a}
$$

The second factor is

$$
C=\sqrt{\frac{1-a^{2}}{a^{2}}}-\frac{1}{a}=\frac{\sqrt{1-a^{2}}-1}{a}
$$

Therefore, our product is

$$
(A+B) C=\frac{\sqrt{1-a^{2}}+1}{a} \cdot \frac{\sqrt{1-a^{2}}-1}{a}=\frac{1-a^{2}-1}{a^{2}}=-1
$$

Győri Matild (Miskolc, Zrínyi Ilona Secondary School II. grade)