Trig 1

Let us take $\frac{\cos2\theta}{1+\sin2\theta} \cdot \frac{1-\sin2\theta}{1-\sin2\theta} = \frac{\cos2\theta(1-\sin2\theta)}{1-\sin^{2}2\theta} = \frac{\cos2\theta(1-\sin2\theta)}{\cos^{2}2\theta} = \sec2\theta - \tan2\theta = \sqrt{\tan^{2}2\theta + 1} - \tan2\theta = \frac{3}{2}.$ Let $x = \tan2\theta$ to give us:

$\sqrt{x^2+1} = x + \frac{3}{2} \Rightarrow x^2 + 1 = x^2 + 3x + \frac{9}{4} \Rightarrow -\frac{5}{4} = 3x \Rightarrow -\frac{5}{12} = \tan2\theta.$

or $\frac{2\tan \theta}{1-\tan^{2}\theta} = -\frac{5}{12};$

or $24\tan \theta = 5\tan^{2}\theta - 5;$

or $0 = 5\tan^{2}\theta - 24\tan \theta - 5;$

or $0 = (5\tan \theta + 1)(\tan \theta - 5)$ ;

or $\tan \theta = -\frac{1}{5}, 5$ . Since $\theta \in (-\frac{\pi}{4},0) \Rightarrow \tan \theta < 0$ . Thus, $\boxed{\tan \theta = -\frac{1}{5}}$ is the only admissible answer.