Given that: $\sin^2 10^{\circ}+\sin^2 20^{\circ}+\sin^2 30^{\circ} \cdots\cdots\cdots \sin^2 80^{\circ}$
we know that:
- $\sin^2\theta+\cos^2\theta=1$
- $\sin(90-\theta)=\cos\theta$
$\therefore \sin^2 10^{\circ}+\sin^2 20^{\circ}+\sin^2 30^{\circ}\cdots \sin^2 (90-30)^{\circ}+\sin^2 (90-20)^{\circ}+\sin^2 (90-10)^{\circ} $
$\implies\sin^2 10^{\circ}+\sin^2 20^{\circ}+\sin^2 30^{\circ}\cdots \cos^2 30^{\circ}+\cos^2 20^{\circ}+\cos^2 10^{\circ}$
$\implies (\sin^210^{\circ}+\cos^210^{\circ})+(\sin^220^{\circ}+\cos^220^{\circ})+(\sin^230^{\circ}+\cos^230^{\circ})+(\sin^240^{\circ}+\cos^240^{\circ})$
$\implies 1+1+1+1=4$
Option (C) is correct.