9 Let $\alpha \in \mathbf{R}$, then the function
$$
f(x)=\frac{\sin (x+\alpha)+\sin (x-\alpha)-4 \sin x}{\cos (x+\alpha)+\cos (x-\alpha)-4 \cos x}, x \in\left(0, \frac{\pi}{4}\right)
$$

the range of the function is
9 Transform the numerator and denominator respectively,
$$
\begin{aligned}
\sin (x+\alpha)+\sin (x-\alpha)-4 \sin x & =2 \sin x(\cos \alpha-2), \\
\cos (x+\alpha)+\cos (x-\alpha)-4 \cos x & =2 \cos x(\cos \alpha-2) .
\end{aligned}
$$

Since $\cos \alpha-2 \neq 0$, then $f(x)=\tan x$, and $x \in\left(0, \frac{\pi}{4}\right)$, therefore $0<$
$$
f(x)<1 \text {. }
$$