# If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ – y cos θ = 0, then prove that x2 + y2 = 1, (where, sin θ ≠ 0 and cos θ ≠ 0).

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x sin θ - y cos θ = 0, (Given)

⇒ x sin θ = y cos θ

⇒ y cos θ = x sin θ

Now dividing both sides by cos θ we get,

y = x ∙ (sin θ/cos θ)

Again, x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x sin3 θ + x ∙ (sin θ /cos θ) ∙ cos3 θ = sin θ cos θ [Since, y = x ∙ (sin θ/cos θ)]

⇒ x sin θ ( sin2 θ + cos2 θ) = sin θ cos θ, [since, cos θ ≠ 0]

⇒ x sin θ (1) = sin θ cos θ,[since, sin2 θ + cos2 θ = 0]

⇒ x sin θ = sin θ cos θ

Now dividing both sides by sin θ we get,

⇒ x = cos θ, [since, sin θ ≠ 0]

Therefore, y = x ∙ (sin θ/cos θ)

⇒ y = cos θ ∙ (sin θ/cos θ), [Putting x = cos θ]

⇒ y = sin θ

Now, x2 + y2

= cos2 θ + sin2 θ

= 1.

Therefore, x2 + y2 = 1.

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