12. Let $f(x)+g(x)=\sqrt{\frac{1+\cos 2 x}{1-\sin x}}\left(x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right)$, and $f(x)$ is an odd function, $g(x)$ is an even function, then $[f(x)]^{2}-[g(x)]^{2}=$ $\qquad$ .
12. $-2 \cos x \quad\left(x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right) \quad$ Hint: $f(-x)+g(-x)=\sqrt{\frac{1+\cos 2 x}{1+\sin x}},-f(x)+g(x)=$ $\sqrt{\frac{1+\cos 2 x}{1+\sin x}}$, combine with the original conditions, solve for $f(x)$ and $g(x)$ and substitute them in.