2. The smallest positive period of the function $y=\frac{7 \sin \tau \tan x}{\sec x \cos 2 x\left(1-\tan ^{2} x\right)}$ is $($.
A. $\frac{1}{4} \pi$
B. $\frac{1}{2} \pi$
C. $\pi$
D. $2 \pi$
2. $\mathrm{C}$

Simplifying $y$ yields $y=\frac{7}{4} \cdot \frac{2 \sin x \cdot 2 \tan x}{\frac{1}{\cos x} \cos 2 x\left(1-\tan ^{2} x\right)}=\frac{7}{4} \cdot \frac{2 \sin x \cos x}{\cos 2 x} \cdot \frac{2 \tan x}{1-\tan ^{2} x}=\frac{7}{4}$ $\tan ^{2} 2 x$

Noting that $\cos x, \cos 2 x$ and $1-\tan ^{2} x \neq 0$, i.e., $x \neq \frac{\pi}{2}+k \pi, \frac{\pi}{4}+\frac{k \pi}{2}$, thus the domain of $y$ is $\left(-\frac{\pi}{2}+k \pi,-\frac{\pi}{4}+k \pi\right) \cup\left(-\frac{\pi}{4}+k \pi, \frac{\pi}{4}+k \pi\right) \cup$ $\left(\frac{\pi}{4}+k \pi, \frac{\pi}{2}+k \pi\right), k \in \boldsymbol{z}$. In summary, the smallest positive period of $y$ is $\pi$.