1. **State the problem:** Prove the trigonometric identity $$\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x.$$\n\n2. **Recall formulas:** Use the double-angle identities: $$\cos 2x = 1 - 2\sin^2 x$$ and $$\sin 2x = 2\sin x \cos x.$$ Also, $$\tan x = \frac{\sin x}{\cos x}.$$\n\n3. **Substitute the double-angle formulas into the left side:**\n$$\frac{1 - (1 - 2\sin^2 x) + 2\sin x \cos x}{1 + (1 - 2\sin^2 x) + 2\sin x \cos x} = \frac{1 - 1 + 2\sin^2 x + 2\sin x \cos x}{1 + 1 - 2\sin^2 x + 2\sin x \cos x} = \frac{2\sin^2 x + 2\sin x \cos x}{2 - 2\sin^2 x + 2\sin x \cos x}.$$\n\n4. **Factor out 2 in numerator and denominator:**\n$$\frac{\cancel{2}(\sin^2 x + \sin x \cos x)}{\cancel{2}(1 - \sin^2 x + \sin x \cos x)} = \frac{\sin^2 x + \sin x \cos x}{1 - \sin^2 x + \sin x \cos x}.$$\n\n5. **Use the Pythagorean identity:** $$1 - \sin^2 x = \cos^2 x,$$ so denominator becomes $$\cos^2 x + \sin x \cos x.$$\n\n6. **Rewrite numerator and denominator:**\nNumerator: $$\sin x (\sin x + \cos x)$$\nDenominator: $$\cos x (\cos x + \sin x).$$\n\n7. **Rewrite the fraction:**\n$$\frac{\sin x (\sin x + \cos x)}{\cos x (\cos x + \sin x)} = \frac{\sin x \cancel{(\sin x + \cos x)}}{\cos x \cancel{(\cos x + \sin x)}} = \frac{\sin x}{\cos x} = \tan x.$$\n\n8. **Conclusion:** The left side simplifies exactly to $$\tan x,$$ proving the identity.