The elimination of θ from xcos θ - ysinθ = 2 and xsinθ + ycosθ

1.	The elimination of θ from xcos θ - ysinθ = 2 and xsinθ + ycosθ = 4 will give:1. x2 - y2 = 202. 3x2 + y2 = 203. x2 + y2 = 204. 3x2 - y2 = 20
Answer» Correct Answer - Option 3 : x² + y² = 20 Given : xcosθ - ysinθ = 2 xsinθ + ycosθ = 4 Formula used: sin²θ + cos²θ = 1 (a + b)² = a² + b² + 2ab Calculation: xcosθ - ysinθ = 2 On squaring we get, ⇒ x²cos²θ + y²sin²θ – 2(xcosθ)(ysinθ) = 4 ….(i) xsinθ + ycosθ = 4 On squaring we get, ⇒ x²sin²θ + y²cos²θ + 2(ycosθ)(xsinθ) = 16 ….(ii) On adding eqns (i) and (ii) ⇒ x²cos²θ + y²sin²θ – 2(xcosθ)(ysinθ) + x²sin²θ + y²cos²θ + 2(ycosθ)(xsinθ) = 4 + 16 ⇒ x²(cos²θ + sin²θ) + y²(sin²θ + cos²θ) = 20 ⇒ x² + y² = 20 ∴ The correct answer is x² + y² = 20

The elimination of θ from xcos θ - ysinθ = 2 and xsinθ + ycosθ = 4 will give:1. x2 - y2 = 202. 3x2 + y2 = 203. x2 + y2 = 204. 3x2 - y2 = 20

Answer» Correct Answer - Option 3 : x² + y² = 20

Given :

xcosθ - ysinθ = 2

xsinθ + ycosθ = 4

Formula used:

sin²θ + cos²θ = 1

(a + b)² = a² + b² + 2ab

Calculation:

xcosθ - ysinθ = 2

On squaring we get,

⇒ x²cos²θ + y²sin²θ – 2(xcosθ)(ysinθ) = 4 ….(i)

xsinθ + ycosθ = 4

On squaring we get,

⇒ x²sin²θ + y²cos²θ + 2(ycosθ)(xsinθ) = 16 ….(ii)

On adding eqns (i) and (ii)

⇒ x²cos²θ + y²sin²θ – 2(xcosθ)(ysinθ) + x²sin²θ + y²cos²θ + 2(ycosθ)(xsinθ) = 4 + 16

⇒ x²(cos²θ + sin²θ) + y²(sin²θ + cos²θ) = 20

⇒ x² + y² = 20

∴ The correct answer is x² + y² = 20