2. Given $\frac{\cos x+\sin x}{\cos x-\sin x}=2007$. Then
$$
\csc \left(\frac{\pi}{2}+2 x\right)+\tan (3 \pi+2 x)
$$

is equal to $(\quad$.
(A) 2005
(B) 2006
(C) 2007
(D) 2008
2.C.
$$
\begin{array}{l}
\text { Given } \frac{\cos x+\sin x}{\cos x-\sin x}=2007, \text { we know that } \\
\frac{1+\tan x}{1-\tan x}=\tan \left(x+\frac{\pi}{4}\right)=2007 . \\
\text { Therefore, } \csc \left(\frac{\pi}{2}+2 x\right)+\tan (3 \pi+2 x) \\
=\frac{1}{\cos 2 x}+\frac{\sin 2 x}{\cos 2 x}=\frac{1+\sin 2 x}{\cos 2 x} \\
=\tan \left(x+\frac{\pi}{4}\right)=2007 .
\end{array}
$$