1. **State the problem:** Prove the expression $$\frac{2\tan x - \sin 2x}{2\sin^2 x}$$ simplifies or equals a certain value. 2. **Rewrite the expression:** Recall that $$\tan x = \frac{\sin x}{\cos x}$$ and $$\sin 2x = 2\sin x \cos x$$. 3. Substitute these into the numerator: $$2\tan x - \sin 2x = 2 \cdot \frac{\sin x}{\cos x} - 2 \sin x \cos x = \frac{2\sin x}{\cos x} - 2 \sin x \cos x$$ 4. Find a common denominator for the numerator terms: $$\frac{2\sin x}{\cos x} - 2 \sin x \cos x = \frac{2\sin x}{\cos x} - \frac{2 \sin x \cos^2 x}{\cos x} = \frac{2\sin x - 2 \sin x \cos^2 x}{\cos x}$$ 5. Factor out $$2\sin x$$ in the numerator: $$\frac{2\sin x (1 - \cos^2 x)}{\cos x}$$ 6. Use the Pythagorean identity $$1 - \cos^2 x = \sin^2 x$$: $$\frac{2\sin x \sin^2 x}{\cos x} = \frac{2 \sin^3 x}{\cos x}$$ 7. Now the entire expression is: $$\frac{\frac{2 \sin^3 x}{\cos x}}{2 \sin^2 x} = \frac{2 \sin^3 x}{\cos x} \cdot \frac{1}{2 \sin^2 x}$$ 8. Simplify by canceling $$2$$ and $$\sin^2 x$$: $$\frac{\sin^3 x}{\cos x} \cdot \frac{1}{\sin^2 x} = \frac{\sin x}{\cos x}$$ 9. Recognize that $$\frac{\sin x}{\cos x} = \tan x$$. **Final answer:** $$\frac{2\tan x - \sin 2x}{2\sin^2 x} = \tan x$$