# If x=a(Sinθ+Cosθ) and y=b(Sin θ-Cosθ), then the value of (x²/a²)+(y²/b²) is:

**Answer: (3) 2
**

**Explanation:-**

X = a(Sinθ + Cosθ)

Square on both side,

X² = a² (Sin²θ + Cos²θ + 2 Sinθ Cosθ)

X²= a² (1 + 2 Sinθ Cosθ)

Y = b (Sinθ – Cosθ)

Square on both side,

Y² = b² (Sin²θ + Cos²θ – 2 Sinθ Cosθ)

Y² = b² (1 – 2 Sinθ Cosθ)

Now, X²/a² + Y²/b²

Put the value of Y² = b² (1 – 2 Sinθ Cosθ) & X²= a² (1 + 2 Sinθ Cosθ)

Then, a² (1 + 2 Sinθ Cosθ)/a² + b² (1 – 2 Sinθ Cosθ)/b²

1 + 2 Sinθ Cosθ + 1 – 2 Sinθ Cosθ

1 + 1

= 2

Hence, the correct answer is option (3) 2.