Correct Answer: $0$

Solution : $\left ( \frac{\sin47^{\circ}}{\cos43^{\circ}} \right )^{2}+\left ( \frac{\cos43^{\circ}}{\sin47^{\circ}} \right )^{2}-4\cos^{2}45^{\circ}$
We know that $\sin (90^{\circ} - \theta) = \cos \theta$ and $\cos (90^{\circ} - \theta) = \sin \theta$
Such that $\sin 47^{\circ} = \cos 43^{\circ}$ and $\cos 43^{\circ} = \sin 47^{\circ}$
$=\left ( \frac{\cos 43^{\circ}}{\cos 43^{\circ}} \right )^{2}+\left ( \frac{\sin 47^{\circ}}{\sin 47^{\circ}} \right )^{2}-4\cos^{2}45^{\circ}$
$=1 + 1 - 4\cos^{2}45^{\circ}$
We know that $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$
$=2 - 4\left(\frac{1}{\sqrt{2}}\right)^{2}$
$= 2 - 4\left(\frac{1}{2}\right) $
$= 2 - 2 $
$= 0$
Hence, the correct answer is $0$.