If x = a (sinθ + cosθ) and y = b(sinθ - cosθ)
then value of \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) is?
Answer & explanation
Correct answer: option 4
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