$$\left(\frac{\sin 47^\circ}{\cos 53^\circ}\right)^2 + \left(\frac{\cos 43^\circ}{\sin 47^\circ}\right)^2 - 4 \cos^2 45^\circ$$

$$\left(\frac{\sin47^{\circ}}{\cos53^{\circ}}\right)^2+\left(\frac{\cos43^{\circ}}{\sin47^{\circ}}\right)^2-4\cos^245^{\circ} = \left(\frac{\sin\left(90-53\right)^{\circ\ }}{\cos53^{\circ\ }}\right)^2+\left(\frac{\cos\left(90-47\right)^{\circ\ }}{\sin47^{\circ\ }}\right)^2-4\left(\frac{1}{\sqrt{2}}\right)^2=\left(\frac{\cos53^{\circ\ }}{\cos53^{\circ\ }}\right)^2+\left(\frac{\sin47^{\circ\ }}{\sin47^{\circ\ }}\right)^2-4\left(\frac{1}{2}\right)=1+1-2=0$$

Hence, the correct answer is Option D