3. In $\triangle A B C$,
$$
\cos A+\sin A-\frac{2}{\cos B+\sin B}=0 \text {. }
$$

Then the value of $\frac{a}{c}$ is $\qquad$
3. $\frac{\sqrt{2}}{2}$.

From the given equation, we have
$$
\sqrt{2} \sin \left(A+\frac{\pi}{4}\right)-\frac{\sqrt{2}}{\sin \left(B+\frac{\pi}{4}\right)}=0,
$$

which means $\sin \left(A+\frac{\pi}{4}\right) \cdot \sin \left(B+\frac{\pi}{4}\right)=1$.
Since $0 \leqslant \sin \left(A+\frac{\pi}{4}\right), \sin \left(B+\frac{\pi}{4}\right) \leqslant 1$, then, $A=B=\frac{\pi}{4}, C=\frac{\pi}{2}$.
By the Law of Sines, we know $\frac{a}{c}=\frac{\sin A}{\sin C}=\frac{\sqrt{2}}{2}$.