Question 76, Given $a, b \in R^{+}, a \sqrt{1-b^{2}}-b \sqrt{1-a^{2}}=a b$, try to find the maximum possible value of $\frac{a}{b}+\frac{b}{a}$.

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Question 76, Solution: According to the conditions, we have:
$$
a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}=\frac{a^{2}\left(1-b^{2}\right)-b^{2}\left(1-a^{2}\right)}{a \sqrt{1-b^{2}}-b \sqrt{1-a^{2}}}=\frac{a^{2}-b^{2}}{a b}
$$

Thus, we know:
$$
\begin{array}{l}
2 a \sqrt{1-b^{2}}=\frac{a^{2}-b^{2}}{a b}+a b \\
\Rightarrow 4 a^{2}\left(1-b^{2}\right)=\frac{\left(a^{2}-b^{2}\right)^{2}}{a^{2} b^{2}}+a^{2} b^{2}+2\left(a^{2}-b^{2}\right)=\frac{\left(a^{2}+b^{2}\right)^{2}}{a^{2} b^{2}}-4+a^{2} b^{2}+2\left(a^{2}-b^{2}\right) \\
\Rightarrow 2\left(a^{2}+b^{2}\right)=\frac{\left(a^{2}+b^{2}\right)^{2}}{a^{2} b^{2}}-4+5 a^{2} b^{2}
\end{array}
$$

Let $s=a b, a+\frac{a}{b}=t$ then $a^{2}+b^{2}=s t$. Substituting into (*) we get:
$$
2 s t=t^{2}-4+5 s^{2}
$$
$\Leftrightarrow 5 s^{2}-2 t \cdot s+\left(t^{2}-4\right)=0$
Considering the above as a quadratic equation in $s$, its discriminant is:
$$
\begin{array}{l}
\Delta=(26)^{2} 74 \times 5 \times\left(t^{2}=4\right) \geq 0 \\
\Rightarrow t \leq \sqrt{5} \\
\text { When } (a, b)=\left(\frac{\sqrt{\frac{5+2 \sqrt{5}}{5}} \sqrt{\frac{5-2 \sqrt{3}}{5}}}{\sqrt{5}}\right. \text {, } \\
\end{array}
$$

the equality holds.