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If $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$ and $x \sin \theta = y \cos \theta$, prove that $x^2 + y^2 = 1$.
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Given, $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta) + y \cos \theta (\cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta) + x \sin \theta (\cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x = \cos \theta$
Given, $x \sin \theta = y \cos \theta$
$\Rightarrow \cos \theta \sin \theta = y \cos \theta$
$\Rightarrow y = \sin \theta$
LHS $= x^2 + y^2 = (\cos \theta)^2 + (\sin \theta)^2 = 1 = \text{RHS}$
$\Rightarrow x \sin \theta (\sin^2 \theta) + y \cos \theta (\cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta) + x \sin \theta (\cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x = \cos \theta$
Given, $x \sin \theta = y \cos \theta$
$\Rightarrow \cos \theta \sin \theta = y \cos \theta$
$\Rightarrow y = \sin \theta$
LHS $= x^2 + y^2 = (\cos \theta)^2 + (\sin \theta)^2 = 1 = \text{RHS}$