2. Given $\theta_{1}, \theta_{2}, \cdots, \theta_{n} \in\left(0, \frac{\pi}{2}\right)$, let
$$
\begin{array}{l}
M=\left(\sum_{i=1}^{\prime \prime} \tan \theta_{i}\right)\left(\sum_{i=1}^{\prime \prime} \cot \theta_{i}\right), \\
N=\left(\sum_{i=1}^{\prime \prime} \sin \theta_{i}\right)\left(\sum_{i=1}^{n} \csc \theta_{i}\right) .
\end{array}
$$

Then the relationship between $M$ and $N$ is ( ).
(A) $M \geqslant N$
(B) $M \leqslant N$
(C) $M=N$
(D) Uncertain
2. A.
$$
\begin{array}{l}
M-N \\
=\left(\sum_{i=1}^{n} \tan \theta_{i}\right)\left(\sum_{i=1}^{n} \cot \theta_{i}\right)-\left(\sum_{i=1}^{n} \sin \theta_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{\sin \theta_{i}}\right) \\
=\frac{1}{2} \sum_{j=1}^{n} \sum_{i=1}^{n}\left(\tan \theta_{i} \cdot \cot \theta_{j}+\tan \theta_{j} \cdot \cot \theta_{i}-\frac{\sin \theta_{i}}{\sin \theta_{j}}-\frac{\sin \theta_{j}}{\sin \theta_{i}}\right) .
\end{array}
$$

Therefore, it is only necessary to prove that each term is non-negative, i.e.,
$$
\tan \theta_{i} \cdot \cot \theta_{j}+\tan \theta_{j} \cdot \cot \theta_{i} \geqslant \frac{\sin \theta_{i}}{\sin \theta_{j}}+\frac{\sin \theta_{j}}{\sin \theta_{i}} .
$$

The above inequality, when brought to a common denominator, becomes
$$
\begin{array}{l}
\left(\sin \theta_{i} \cdot \cos \theta_{j}\right)^{2}+\left(\sin \theta_{j} \cdot \cos \theta_{i}\right)^{2} \\
\geqslant\left(\sin ^{2} \theta_{i}+\sin ^{2} \theta_{j}\right) \cos \theta_{i} \cdot \cos \theta_{j} \\
\Leftrightarrow \sin ^{2} \theta_{i}+\sin ^{2} \theta_{j}-2 \sin ^{2} \theta_{i} \cdot \sin ^{2} \theta_{j} \\
\quad \geqslant\left(\sin ^{2} \theta_{i}+\sin ^{2} \theta_{j}\right) \cos \theta_{i} \cdot \cos \theta_{j} \\
\Leftrightarrow\left(\sin ^{2} \theta_{i}+\sin ^{2} \theta_{j}\right)\left(1-\cos \theta_{i} \cdot \cos \theta_{j}\right) \\
\quad \geqslant 2 \sin ^{2} \theta_{i} \cdot \sin ^{2} \theta_{j} .
\end{array}
$$

By $\sin ^{2} \theta_{i}+\sin ^{2} \theta_{j} \geqslant 2 \sin \theta_{i} \cdot \sin \theta_{j}$ and
$$
\begin{array}{l}
1-\cos \theta_{i} \cdot \cos \theta_{j}-\sin \theta_{i} \cdot \sin \theta_{j} \\
=1-\cos \left(\theta_{i}-\theta_{j}\right) \geqslant 0
\end{array}
$$

we know that $M \geqslant N$ holds.