3. Satisfy
$$
\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}+\frac{1-\sin \theta-\cos \theta}{1-\sin \theta+\cos \theta}=2
$$

The maximum negative angle $\theta$ in radians is
3. $-\frac{\pi}{2}$.

Given the equation can be transformed into
$$
\frac{(1-\sin \theta+\cos \theta)^{2}+(1-\sin \theta-\cos \theta)^{2}}{(1-\sin \theta)^{2}-\cos ^{2} \theta}=2 \text {. }
$$

Simplifying yields $\cos ^{2} \theta=0 \Rightarrow \cos \theta=0$.
Thus, the largest negative angle $\theta$ that satisfies the condition is $\theta=-\frac{\pi}{2}$.