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x sin3 θ + y cos3 θ = sin θ cos θ -> eq 1 x sin θ − y cos θ = 0 x sin θ = y cos θ -->eq2 substituting in eq1 y cos θ sin2 θ + y cos3 θ = sin θ cos θ taking y cos θ common y cos θ(sin2 θ + cos2 θ) = sin θ cos θ { we know sin2 θ + cos2 θ =1} y cos θ = sin θ cos θ y = sin θ substituting in eq 2 x sin θ = sin θ cos θ x = cos θ x2 + y2 sin2 θ + cos2 θ = 1
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