一道三角不等式题的证明

已知$0<\theta_1<\theta_2<\dots<\theta_n<\frac\pi2$，求证：

$\cot\theta_n<\frac{\sum^n_{i=1}\cos\theta_i}{\sum^n_{i=1}\sin\theta_i}<\cot\theta_1$

（命题$946$）

$0<\theta_1<\theta_2<\dots<\theta_n<\frac\pi2$

$\Rightarrow\left\{ \begin{aligned} \sin\theta_i&>&0(i=1,2,\dots,n) \\ \cot\theta_1&>&\cot\theta_2>\dots>\cot\theta_n \end{aligned} \right.$

$\cot\theta_1>\cot\theta_i>\cot\theta_n(i=2,3,\dots,n-1)$

$\Rightarrow\cot\theta_1\sin\theta_i>\cos\theta_i>\cot\theta_n\sin\theta_i$

$\Rightarrow\cot\theta_1\cdot\sum^n_{i=1}\sin\theta_i>\sum^n_{i=1}\cos\theta_i>\cot\theta_n\cdot\sum^n_{i=1}\sin\theta_i$

$\sum^n_{i=1}\sin\theta_i>0\Rightarrow\cot\theta_n<\frac{\sum^n_{i=1}\cos\theta_i}{\sum^n_{i=1}\sin\theta_i}<\cot\theta_1$