(3) The maximum negative angle $\theta$ in radians that satisfies $\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}+\frac{1-\sin \theta-\cos \theta}{1-\sin \theta+\cos \theta}=2$ is $\qquad$.
(3) $-\frac{\pi}{2}$ Hint: Method 1 The given equation can be transformed into
$$
\frac{(1-\sin \theta+\cos \theta)^{2}+(1-\sin \theta-\cos \theta)^{2}}{(1-\sin \theta)^{2}-\cos ^{2} \theta}=2 .
$$

Simplifying yields $\cos ^{2} \theta=0$, i.e., $\cos \theta=0$.
Therefore, the maximum negative angle $\theta=-\frac{\pi}{2}$.
Method 2 $2=\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}+\frac{1-\sin \theta-\cos \theta}{1-\sin \theta+\cos \theta}$
$$
\geqslant 2 \sqrt{\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta} \cdot \frac{1-\sin \theta-\cos \theta}{1-\sin \theta+\cos \theta}}=2 \text {, }
$$

with equality holding if and only if
$$
\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=\frac{1-\sin \theta-\cos \theta}{1-\sin \theta+\cos \theta}=1,
$$

i.e., $\cos \theta=0$.
At this point, the maximum negative angle $\theta=-\frac{\pi}{2}$.