If x*sin^3 θ + y *cos^3 θ = sinθ * cosθ ≠ 0 and x sinθ - y cosθ = 0, then value of (x^2 + y^2 ) is

Given, xsinθ -ycosθ = 0
y=xsinθcosθ⇒y=xsinθcosθ
Given the expression:
xsin$$^3$$θ + ycos$$^3$$θ = sinθcosθ
Replacing the value of y we get,
xsin$$^3$$θ+(xsinθ/cosθ)×cos$$^3$$θ=sinθcosθ
x sin3θ + x sinθcos$$^2$$θ = sinθcosθ
x sinθ × (sin2θ + cos2θ) = sinθcosθ
We know the identity: sin2θ + cos2θ = 1
x = cosθ
y=xsinθ/cosθ
y = sinθ
Hence,
x$$^2$$ + y$$^2$$ = cos$$^2$$θ + sin$$^2$$θ
= 1
Option A is the correct answer.