1. The range of the function $f(x)=\sin ^{4} x \tan x+\cos ^{4} x \cot x$ is
Solution: $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2},+\infty\right)$
Notice that
$$
f(x)=\sin ^{4} x \frac{\sin x}{\cos x}+\cos ^{4} x \frac{\cos x}{\sin x}=\frac{\sin ^{6} x+\cos ^{6} x}{\sin x \cos x}=\frac{2-\frac{3}{2} \sin ^{2} 2 x}{\sin 2 x}
$$

Let $t=\sin 2 x \in[-1,0) \cup(0,1]$, then
$$
f(x)=\frac{2-\frac{3}{2} t^{2}}{t}=\frac{2}{t}-\frac{3}{2} t
$$

It is easy to see that $g(t)=\frac{2}{t}-\frac{3}{2} t$ is a decreasing function on the intervals $[-1,0)$ and $(0,1]$.
Therefore, the range of $f(x)$ (i.e., $g(t)$) is $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2},+\infty\right)$.