1. If $0<a<1$, then $\sqrt{a^{2}+\frac{1}{a^{2}}-2} \div\left(1+\frac{1}{a}\right) \times$ $\frac{1}{1+a}$ can be simplified to ( ).
(A) $\frac{1-a}{1+a}$
(B) $\frac{a-1}{a+1}$
(C) $1-a^{2}$
(D) $a^{2}-1$
$\begin{array}{l}\text { 1. } \because \sqrt{\left(a-\frac{1}{a}\right)^{2}}=\left(\frac{1}{a}-a\right)=\frac{1-a^{2}}{a}, \\ \therefore \text { original expression }=\frac{1-a^{2}}{a} \times \frac{a}{a+1} \times \frac{1}{1+a}=\frac{1-a}{1+a} .\end{array}$