If x = a (sinθ + cosθ) and y = b(sinθ - cosθ)

then value of \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) is?

2

x = a (sinθ + cosθ)

⇒ (\(\frac{x}{a}\))²  =  (sinθ + cosθ)²

⇒ \(\frac{x^2}{a^2}\) = 1 + 2sinθcosθ               (because sin²θ + cos²θ = 1)

y = b (sinθ - cosθ)

⇒ (\(\frac{y}{b}\))² = (sinθ - cosθ)²

⇒ \(\frac{y^2}{b^2}\) = 1 - 2sinθcosθ

Thus ; \(\frac{x^2}{a^2}\)  + \(\frac{y^2}{b^2}\) = (1+2sinθcosθ) + (1-2sinθcosθ) = 2