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Best answer

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.