Given that $tan \alpha = -\frac{1}{2}$, find the value of $\frac{2 \sin \alpha \cos \alpha}{\sin^2 \alpha - \cos^2 \alpha}$.

1. Problem Description

Given that $tan \alpha = -\frac{1}{2}$, find the value of $\frac{2 \sin \alpha \cos \alpha}{\sin^2 \alpha - \cos^2 \alpha}$.

2. Solution Steps

First, divide both the numerator and the denominator of the given expression by $\cos^2 \alpha$:

$$\frac{2 \sin \alpha \cos \alpha}{\sin^2 \alpha - \cos^2 \alpha} = \frac{\frac{2 \sin \alpha \cos \alpha}{\cos^2 \alpha}}{\frac{\sin^2 \alpha - \cos^2 \alpha}{\cos^2 \alpha}} = \frac{2 \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin^2 \alpha}{\cos^2 \alpha} - \frac{\cos^2 \alpha}{\cos^2 \alpha}} = \frac{2 \tan \alpha}{\tan^2 \alpha - 1}$$

Now, substitute the given value $tan \alpha = -\frac{1}{2}$ into the simplified expression:

$$\frac{2 \tan \alpha}{\tan^2 \alpha - 1} = \frac{2 \cdot (-\frac{1}{2})}{(-\frac{1}{2})^2 - 1} = \frac{-1}{\frac{1}{4} - 1} = \frac{-1}{\frac{1}{4} - \frac{4}{4}} = \frac{-1}{-\frac{3}{4}} = \frac{-1}{1} \cdot \frac{4}{-3} = \frac{4}{3}$$

3. Final Answer

The value of $\frac{2 \sin \alpha \cos \alpha}{\sin^2 \alpha - \cos^2 \alpha}$ is $\frac{4}{3}$.