CUET Preparation Today
If x sin3θ + y cos3θ = sinθcosθ, and x sinθ - y cosθ = 0, then find the value of (x2 + y2). |
sinθ - cosθ sinθ + cosθ sinθ.cosθ 1 |
1 |
x sin3θ + y cos3θ = sinθcosθ ----(i) & x sinθ - y cosθ = 0 ⇒ x sinθ = ycosθ ----- (ii) in eqn (i) y cosθ.sin2θ + ycosθ.cos2θ = sinθcosθ y cosθ (sin2θ + cos2θ) = sinθcosθ y = sinθ from eqn (ii) x = cosθ ∴ x2 + y2 = sin2θ + cos2θ = 1 |
