1. **State the problem:** Simplify the expression $$\frac{1-2\sin x \cos x}{1+2\sin x \cos x}$$. 2. **Recall the double-angle identity:** We know that $$\sin(2x) = 2\sin x \cos x$$. 3. **Substitute the identity:** Replace $$2\sin x \cos x$$ with $$\sin(2x)$$ in numerator and denominator: $$\frac{1 - \sin(2x)}{1 + \sin(2x)}$$. 4. **Rewrite the expression:** The expression is now $$\frac{1 - \sin(2x)}{1 + \sin(2x)}$$. 5. **Use the tangent half-angle substitution:** Let $$t = \tan x$$, but a more straightforward approach is to multiply numerator and denominator by the conjugate of the denominator: Multiply numerator and denominator by $$1 - \sin(2x)$$: $$\frac{(1 - \sin(2x))^2}{(1 + \sin(2x))(1 - \sin(2x))} = \frac{(1 - \sin(2x))^2}{1 - \sin^2(2x)}$$. 6. **Simplify denominator using Pythagorean identity:** $$1 - \sin^2(2x) = \cos^2(2x)$$. 7. **Rewrite the expression:** $$\frac{(1 - \sin(2x))^2}{\cos^2(2x)} = \left(\frac{1 - \sin(2x)}{\cos(2x)}\right)^2$$. 8. **Express numerator in terms of cosine:** Recall that $$1 - \sin(2x) = \cos(2x) \tan\left(\frac{\pi}{4} - x\right)$$, but a simpler way is to write: $$\frac{1 - \sin(2x)}{\cos(2x)} = \frac{\cos(2x) - \sin(2x)}{\cos(2x)} = 1 - \tan(2x)$$. 9. **Final simplified form:** $$\left(1 - \tan(2x)\right)^2$$. **Answer:** $$\boxed{\left(1 - \tan(2x)\right)^2}$$