1. Simplify
$$
\sqrt{1+2 \sin \alpha \cdot \cos \alpha}+\sqrt{1-2 \sin \alpha \cdot \cos \alpha}
$$
$\left(0^{\circ}<\alpha \leqslant 90^{\circ}\right)$ The result is $\qquad$ .
II. 1.2 $2 \cos \alpha$ or $2 \sin \alpha$.
It is easy to know that $1+2 \sin \alpha \cdot \cos \alpha$
$$
\begin{array}{l}
=\sin ^{2} \alpha+2 \sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha \\
=(\sin \alpha+\cos \alpha)^{2} .
\end{array}
$$

Similarly, $1-2 \sin \alpha \cdot \cos \alpha=(\sin \alpha-\cos \alpha)^{2}$. Therefore, the original expression is
$$
\begin{array}{l}
=\sqrt{(\sin \alpha+\cos \alpha)^{2}}+\sqrt{(\sin \alpha-\cos \alpha)^{2}} \\
=|\sin \alpha+\cos \alpha|+|\sin \alpha-\cos \alpha| .
\end{array}
$$

Since $0^{\circ}<\alpha<90^{\circ}$, we have $\sin \alpha>0, \cos \alpha>0$, and when $0^{\circ}<\alpha<45^{\circ}$, $\sin \alpha<\cos \alpha$;
when $45^{\circ} \leqslant \alpha \leqslant 90^{\circ}$, $\sin \alpha \geqslant \cos \alpha$.
Thus, (1) when $0^{\circ}<\alpha<45^{\circ}$,
the original expression is $(\sin \alpha+\cos \alpha)+(\cos \alpha-\sin \alpha)$
$$
=2 \cos \alpha \text {; }
$$
(2) when $45^{\circ} \leqslant \alpha \leqslant 90^{\circ}$,
$$
\begin{array}{l}
\text { the original expression }=(\sin \alpha+\cos \alpha)+(\sin \alpha-\cos \alpha) \\
=2 \sin \alpha .
\end{array}
$$