1. **State the problem:** Prove that $$\frac{1 - \sin^{2} x}{\cos x} = \frac{\sin 2x}{2 \sin x}$$. 2. **Recall the Pythagorean identity:** $$1 - \sin^{2} x = \cos^{2} x$$. 3. **Rewrite the left side (LHS):** $$\frac{1 - \sin^{2} x}{\cos x} = \frac{\cos^{2} x}{\cos x}$$ 4. **Simplify the fraction by canceling one $\cos x$ factor:** $$\frac{\cancel{\cos x} \cdot \cos x}{\cancel{\cos x}} = \cos x$$ 5. **Rewrite the right side (RHS):** Recall the double angle formula for sine: $$\sin 2x = 2 \sin x \cos x$$ So, $$\frac{\sin 2x}{2 \sin x} = \frac{2 \sin x \cos x}{2 \sin x}$$ 6. **Simplify the RHS by canceling $2 \sin x$:** $$\frac{\cancel{2} \cancel{\sin x} \cos x}{\cancel{2} \cancel{\sin x}} = \cos x$$ 7. **Compare LHS and RHS:** Both sides equal $$\cos x$$. **Therefore, the identity is proven:** $$\frac{1 - \sin^{2} x}{\cos x} = \frac{\sin 2x}{2 \sin x}$$.