1. Let functions $f(x), g(x)$ satisfy $f(x)+g(x)=\sqrt{\frac{1+\cos 2 x}{1-\sin x}}\left(x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right)$, and $f(x)$ is an odd function, $g(x)$ is an even function, then $(f(x))^{2}-(g(x))^{2}=$ $\qquad$
1. $-2 \cos x$.

Since
$$
-f(x)+g(x)=f(-x)+g(-x)=\sqrt{\frac{1+\cos 2 x}{1+\sin x}},
$$

therefore
$$
\begin{aligned}
(f(x))^{2}-(g(x))^{2} & =(f(x)+g(x)) \cdot(f(x)-g(x))=\sqrt{\frac{1+\cos 2 x}{1-\sin x}} \cdot\left(-\sqrt{\frac{1+\cos 2 x}{1+\sin x}}\right) \\
& =-\frac{1+\cos 2 x}{\cos x}=-2 \cos x .
\end{aligned}
$$