Range of $\frac{\cos\theta_1+\cdots+\cos\theta_{10}}{\sin\theta_1+\cdots+\sin\theta_{10}}$ given $\sin^2\theta_1+\cdots+\sin^2\theta_{10}=1$ 
Given that for $ \theta_i \in \left[0, \dfrac{\pi}{2}\right]$, where $1 \le i \le 10$ , $\sin^2\theta_1+\sin^2\theta_2+\cdots+\sin^2\theta_{10}=1$, find the minimum and maximum value of $$\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}}$$

I tried using Cauchy Schwarz inequality which just gives $\displaystyle \sum_{i=1}^{10} \sin\theta_i \le \sqrt{10}$ and $\displaystyle \sum_{i=1}^{10} \cos \theta_i \le \sqrt{90}$ because $\displaystyle \sum_{i=1}^{10}\cos^2\theta_i =9$ but I don't think it would be helpful here.
Maybe some inequality can be applied by writing it as $$\dfrac{\displaystyle \sum_{i=1}^{10} \sqrt{1-\sin^2\theta_i}}{\displaystyle \sum_{i=1}^{10} \sin \theta_i}$$
Any hints would be appreciated!

The method suggested by @Albus Dumbledore can be tweaked slightly to prove that the upper bound of the given expression is $9$. Note that, since $\theta_i \in \left[0,\dfrac{\pi}{2}\right]$, $\cos(\theta_i) \geq 0$ and $\sin(\theta_i) \geq 0$. We will show that $\cos(\theta_i) \leq S_i$:
\begin{align}
\cos(\theta_i) \leq S_i & \iff \cos^2(\theta_i) \leq S_i^2 \\
& \iff \sum_{j \neq i} \sin^2(\theta_j) \leq \left( \sum_{j \neq i}  \sin(\theta_j) \right)^2
\end{align}
Which is obviously true.
Thus, $\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}} \leq  \frac{S_1+S_2+S_3+\cdots+S_{10}}{\sin \theta_1+\sin \theta_2+\cdots+\sin \theta_{10}}=\boxed{9}$ . Equality holds when $\theta_1=\dfrac{\pi}{2}, \theta_2=...=\theta_n=0$, up to permutation.
