For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
(i) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
(ii) there exists $x \in\left(1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right)$ such that
$$
\begin{array}{l}
{\left[(1-x) \sin \theta-\sqrt{a-x^{2} \cos ^{2} \theta}\right]^{2}+\lfloor x \cos \theta} \\
\sqrt{\left.a-(1-x)^{2} \sin ^{2} \theta\right]^{2}} \leqslant a .
\end{array}
$$
$$
\begin{array}{l}
\left.x \sqrt{\frac{a}{\sin ^{2} \theta}-(1-x)^{2}}\right) \geqslant a \text { . } \\
\text { First, prove a lemma: Let } 0 \\
1, p^{2}+q^{2} \leqslant 1, f(x)=(1-x) \sqrt{p^{2} \cdots x^{2}}+ \\
x \sqrt{q^{2}-(1-x)^{2}} \quad(1-q \leqslant x \leqslant p) \text {. } \\
\text { When } \sqrt{p^{2}-x^{2}}=\sqrt{q^{2}-(1-x)^{2}}, \text { i.e., } x= \\
\frac{p^{2}+1}{2} \in(1-q, p), \text { then } f(x) \text { reaches its maximum value. } \\
\end{array}
$$

Solution: From (i),
(1)
(ii) is equivalent to: there exists $x \in\left(1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right)$, satisfying

Proof of the lemma:
By. 7: $1-q \leqslant x \leqslant p$, thus we can let
$x=p \sin \alpha, 1, x=q \sin \beta, 00,
$$
The roots of the left side of (4) are
$$
\begin{array}{l}
\frac{\sin ^{4} \theta \cos ^{4} \theta}{1-3 \sin ^{2} \theta \cos ^{2} \theta}\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta} \pm .\right. \\
\left.\sqrt{\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\right)^{2}-\frac{1}{\cos ^{4} \theta}-\frac{1}{\sin ^{4} \theta}+\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}}\right) \\
=\frac{\sin ^{2} \theta \cos ^{2} \theta}{1 \mp \sqrt{3} \sin \theta \cos \theta},
\end{array}
$$

Therefore, from (4) we get
$\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} \leqslant a \leqslant \frac{\sin ^{2} \theta \cos ^{2} \theta}{1-\sqrt{3} \sin \theta \cos \theta}$.

Since $\frac{\sin ^{2} \theta \cos ^{2} \theta}{(\sin \theta+\cos \theta)^{2}}<\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta}$,
Therefore, when
$$
\begin{array}{l}
a=\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} \text {, it satisfies (1). Hence, the required } \\
a=\frac{\sin ^{2} \theta \cos \theta}{1+\sqrt{3} \sin \theta \cos \theta} .
\end{array}
$$