Range of $\frac{\cos\theta_1+\cdots+\cos\theta_{10}}{\sin\theta_1+\cdots+\sin\theta_{10}}$ given $\sin^2\theta_1+\cdots+\sin^2\theta_{10}=1$

lower bound:

Notice that $$\cos ^2 \theta_1=\sin^2 \theta_2+\sin^2 \theta_3+\cdots+\sin^2 \theta_{10}$$ similarly $$\cos^2 \theta_i=\sum_{j=1 \rightarrow 10 , j\neq i }\sin^2 \theta_j$$
$$S_i=\sum_{j=1\rightarrow 10,j\neq i} \sin \theta_i \tag {say}$$ thus by power mean inequality

$$\cos \theta_i\ge \sqrt{\frac{S_i^2}{9}}=\frac{S_i}{3}$$ Hence $$\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}}\ge \frac{S_1+S_2+S_3+\cdots+S_{10}}{3(\sin \theta_i+\sin \theta_2+\cdots+\sin \theta_{10})}=\boxed 3$$

Upper bound: see See Hai's beautiful proof!

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.