1. **State the problem:** Given $x = \cos \theta - i \sin \theta$, find the value of $x - \frac{1}{x}$. 2. **Recall the expression for $x$:** We have $x = \cos \theta - i \sin \theta$. 3. **Find $\frac{1}{x}$:** Since $x = \cos \theta - i \sin \theta$, its reciprocal is $$\frac{1}{x} = \frac{1}{\cos \theta - i \sin \theta}.$$ Multiply numerator and denominator by the complex conjugate $\cos \theta + i \sin \theta$ to rationalize the denominator: $$\frac{1}{x} = \frac{\cos \theta + i \sin \theta}{(\cos \theta)^2 + (\sin \theta)^2} = \cos \theta + i \sin \theta,$$ since $\cos^2 \theta + \sin^2 \theta = 1$. 4. **Calculate $x - \frac{1}{x}$:** Substitute the expressions: $$x - \frac{1}{x} = (\cos \theta - i \sin \theta) - (\cos \theta + i \sin \theta) = \cos \theta - i \sin \theta - \cos \theta - i \sin \theta = -2i \sin \theta.$$ 5. **Final answer:** $$x - \frac{1}{x} = -2i \sin \theta.$$ This matches the first option, $-2i \sin \theta$.