Answer
$-\sqrt {sec^2\theta-1}$
Work Step by Step
As $\theta$ is in Quadrant II, we have $sin\theta\gt0, cos\theta\lt0,tan\theta\lt0$
so $tan\theta=\frac{sin\theta}{cos\theta}=\frac{\sqrt {1-cos^2\theta}}{cos\theta}$
use the relationship $cos\theta=\frac{1}{sec\theta}$ we get
$tan\theta=\frac{\sqrt {1-(\frac{1}{sec\theta})^2}}{\frac{1}{sec\theta}}=-\sqrt {sec^2\theta-1}$
