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If $x = 2010$ $\sin θ$ and $y = 2010 \cos θ$ , then $x^{2} + y^{2} =$

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$2010$

${(2010)}^{2}$

$4020$

answer is B.

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Detailed Solution

$\sin^{2} θ + \cos^{2} θ = 1$

$x^{2} + y^{2} = {(2010 \sin θ)}^{2} + {(2010 \cos θ)}^{2} = 2010^{2} \sin^{2} θ + 2010^{2} \cos^{2} θ$

$= 2010^{2} (\sin^{2} θ + \cos^{2} θ) = 2010^{2}$

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