Answer
$\tan\theta$
Work Step by Step
$\frac{x}{\sqrt{4-x^2}}$
Substituting $x=2\sin \theta$ as given in the problem:
$=\frac{2\sin\theta}{\sqrt{4-(2\sin\theta)^2}}$
$=\frac{2\sin\theta}{\sqrt{4-4\sin^2\theta}}$
$=\frac{2\sin\theta}{\sqrt{4(1-\sin^2\theta)}}$
$=\frac{2\sin\theta}{\sqrt{4\cos^2\theta}}$
(Since it's given in the problem that $-\pi/2 < \theta < -\pi/2$, $\cos \theta$ is positive, so $\sqrt{4\cos^2\theta}=2\cos\theta$.)
$=\frac{2\sin\theta}{2\cos\theta}$
$=\tan \theta$
