5. Let $\angle A, \angle B, \angle C$ be the three interior angles of $\triangle ABC$. If $\sin A=a, \cos B=b$, where $a>0, b>0$, and $a^{2}+b^{2} \leqslant 1$, then $\tan C=$ $\qquad$
5. $\frac{a b+\sqrt{1-a^{2}} \sqrt{1-b^{2}}}{a \sqrt{1-b^{2}}-b \sqrt{1-a^{2}}}$.

Since $\cos B=b>0$, we have
$\angle B$ is an acute angle,
$\sin B=\sqrt{1-\cos ^{2} B}=\sqrt{1-b^{2}}$.
Also, $a^{2}+b^{2} \leqslant 1$, then
$\sin A=a \leqslant \sqrt{1-b^{2}}=\sin B$.
Thus, $\sin (\pi-A) \leqslant \sin B$.
If $\angle A$ is an obtuse angle, then $\pi-\angle A$ is an acute angle.
Since $\angle B$ is an acute angle, then
$$
\pi-\angle A \leqslant \angle B \Rightarrow \angle A+\angle B \geqslant \pi \text {, }
$$

which is a contradiction.
Therefore, $\angle A$ is an acute angle, and
$$
\cos A=\sqrt{1-\sin ^{2} A}=\sqrt{1-a^{2}} .
$$

Hence, $\tan A=\frac{\sin A}{\cos A}=\frac{a}{\sqrt{1-a^{2}}}$,
$$
\tan B=\frac{\sin B}{\cos B}=\frac{\sqrt{1-b^{2}}}{b} \text {. }
$$

Then $\tan C=\frac{\tan A+\tan B}{\tan A \cdot \tan B-1}$
$$
=\frac{a b+\sqrt{1-a^{2}} \sqrt{1-b^{2}}}{a \sqrt{1-b^{2}}-b \sqrt{1-a^{2}}} .
$$