Correct Answer: 2

Solution : We know: $\sin (90^{\circ} -A) = \cos A$ and $\sin^2 A+\cos^2 A = 1$
$\frac{\sin^2 32^{\circ}+\sin^2 58^{\circ}}{\cos^2 32^{\circ}+\cos^2 58^{\circ}}+\sin^2 53^{\circ}+\cos 53^{\circ} \sin 37^{\circ}$
= $\frac{\sin^2 32^{\circ}+\sin^2 (90-32)^{\circ}}{\cos^2 32^{\circ}+\cos ^2 (90-32)^{\circ}}+\sin^2 53^{\circ}+\cos 53^{\circ} \sin (90-53)^{\circ}$
= $\frac{\sin^2 32^{\circ}+\cos^2 32^{\circ}}{\cos^2 32^{\circ}+\sin^2 32^{\circ}}+\sin^2 53^{\circ}+\cos 53^{\circ} \cos 53^{\circ}$
= $1+ \sin^2 53^{\circ}+\cos^2 53^{\circ}$
= $1+ 1$
= $2$
Hence, the correct answer is 2.