sin ^{2} 1^{circ} + sin ^{2} 3^{circ} + sin ^ | vedclass

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(C) We are given the expression: $\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ}$.Using the trigonometric ide

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(C) We are given the expression: $\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ}$.Using the trigonometric identity $\sin(90^{\circ} - 	heta) = \cos 	heta$,we can rewrite the terms:$\sin ^{2} 87^{\circ} = \sin ^{2}(90^{\circ} - 3^{\circ}) = \cos ^{2} 3^{\circ}$$\sin ^{2} 89^{\circ} = \sin ^{2}(90^{\circ} - 1^{\circ}) = \cos ^{2} 1^{\circ}$Substituting these into the original expression:$= \sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ} + \cos ^{2} 1^{\circ}$Grouping the terms with the same angle:$= (\sin ^{2} 1^{\circ} + \cos ^{2} 1^{\circ}) + (\sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ})$Using the identity $\sin ^{2} 	heta + \cos ^{2} 	heta = 1$:$= 1 + 1 = 2$.

$\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ} = \ldots$

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