7. The function $y=\tan \omega x(\omega>0)$ intersects the line $y=a$ at points $A, B$, and the minimum value of $|A B|$ is $\pi$, then the interval of monotonic increase for the function $f(x)=\sqrt{3} \sin \omega x-\cos \omega x$ is
A. $\left[2 k \pi-\frac{\pi}{6}, 2 k \pi+\frac{\pi}{6}\right](k \in \mathbf{Z})$
B. $\left[2 k \pi-\frac{\pi}{3}, 2 k \pi+\frac{2 \pi}{3}\right](k \in \mathbf{Z})$
C. $\left[2 k \pi-\frac{2 \pi}{3}, 2 k \pi+\frac{\pi}{3}\right](k \in \mathbf{Z})$
D. $\left[2 k \pi-\frac{\pi}{6}, 2 k \pi+\frac{5 \pi}{6}\right](k \in \mathbf{Z})$
The function $y=\tan \omega x$ has the smallest positive period $T=\pi \Rightarrow \omega=1$, then $f(x)=\sqrt{3} \sin x-\cos x=2 \sin \left(x-\frac{\pi}{6}\right)$, $x-\frac{\pi}{6} \in\left[2 k \pi-\frac{\pi}{2}, 2 k \pi+\frac{\pi}{2}\right] \Rightarrow x \in\left[2 k \pi-\frac{\pi}{3}, 2 k \pi+\frac{2 \pi}{3}\right]$, so the answer is $B$.