10. The solution set of the inequality system $\sin x>\cos x>\tan x>\cot x$ in $(0,2 \pi)$ (expressed in intervals) is $\qquad$
10. $\left(\frac{3 \pi}{4}, \pi-\arcsin \frac{\sqrt{5}-1}{2}\right)$.

As shown in Figure 1, solve using the interval method on the quadrant diagram.
The boundary lines are $\frac{k \pi}{4}(0 \leqslant k \leqslant 7)$ and the solutions to the equation $\cos x = \tan x$ are
$$
\begin{array}{l}
\arcsin \frac{\sqrt{5}-1}{2}, \\
\pi-\arcsin \frac{\sqrt{5}-1}{2} .
\end{array}
$$

By $\sin x > \cos x$, exclude interval (1);
By $\tan x > \cot x$, exclude interval (2);
By $\cos x > \tan x$, exclude interval (3).
Therefore, the solution set is $\left(\frac{3 \pi}{4}, \pi-\arcsin \frac{\sqrt{5}-1}{2}\right)$.