**Q)** If x sin^{3} θ + y cos^{3} θ = sin θ cos θ and x sin θ = y cos θ, prove that x^{2} + y^{2} = 1

**Ans: **Given that x sin^{3} θ + y cos^{3} θ = sin θ cos θ

∴ x sin θ sin^{2} θ + y cos θ cos^{2} θ = sin θ cos θ …….. (i)

**Step 1:** since we are given that x sin θ = y cos θ, let’s substitute the same in the equation (i), we get:

∴ x sin θ sin^{2} θ + (x sin θ) cos^{2} θ = sin θ cos θ

∴ x sin θ (sin^{2} θ + cos^{2} θ) = sin θ cos θ

∴ x sin θ = sin θ cos θ [ since sin^{2} θ + cos^{2} θ = 1]

∴ x = cos θ

**Step 2:** Since, we had x sin θ = y cos θ

by substituting x = cos θ in the above, we get:

x sin θ = y (x)

∴ y = sin θ

**Step 3:** let’s find the value of x ^{2} + y ^{2}

We just calculated x = cos θ and y = sin θ

∴ x ^{2} + y ^{2 }= (cos θ) ^{2} + (sin θ) ^{2}

= cos^{2} θ + sin^{2} θ

= 1

**∴ x ^{2} + y ^{2 }= 1**

**Hence Proved !**

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