1. Datos del problema:

• Numerador: $(1 - \sin 2x)(\cos 2x - \sin 2x - 1)$.
• Denominador: $\sin 2x \sin(\pi/4 - x) \sin(\pi/4 + x)$.

2. Fórmulas/Propiedades:

• $\sin(A-B)\sin(A+B) = \sin^2 A - \sin^2 B$.
• $1 - \sin 2x = (\cos x - \sin x)^2$.
• $\cos 2x - 1 = -2 \sin^2 x $.

3. Desarrollo paso a paso:
Simplificamos el producto de senos en el denominador ($ D $):
$$ \sin(\pi/4 - x) \sin(\pi/4 + x) = \sin^2(\pi/4) - \sin^2 x = \frac{1}{2} - \sin^2 x = \frac{1 - 2 \sin^2 x}{2} = \frac{\cos 2x}{2} $$
Por lo tanto, $D = \sin 2x \cdot \frac{\cos 2x}{2} = \frac{1}{2} \sin 2x \cos 2x = \frac{1}{4} \sin 4x $.

Simplificamos el numerador ($ N $):
$$ N = (1 - \sin 2x) [(\cos 2x - 1) - \sin 2x] = (1 - \sin 2x) [-2 \sin^2 x - 2 \sin x \cos x] $$
$$ N = (1 - \sin 2x) [-2 \sin x (\sin x + \cos x)] $$
Sustituyendo $1 - \sin 2x = (\cos x - \sin x)^2$:
$$ N = (\cos x - \sin x)^2 [-2 \sin x (\cos x + \sin x)] = -2 \sin x (\cos x - \sin x) (\cos^2 x - \sin^2 x) $$
$$ N = -2 \sin x (\cos x - \sin x) \cos 2x $$

Calculamos $H$:
$$ H = \frac{-2 \sin x (\cos x - \sin x) \cos 2x}{\frac{1}{2} \sin 2x \cos 2x} = \frac{-4 \sin x (\cos x - \sin x)}{2 \sin x \cos x} = -2 \frac{\cos x - \sin x}{\cos x} $$
$$ H = -2(1 - \tan x) = 2 \tan x - 2 $$

4. Resultado final:
$$ H = \tan x - 1 $$