Question

If x = a sin theta +b cos theta and y= a cos theta +bsin theta, prove that x² + y square = a square +b square​

Answer 1

Correct question :-

If x = a sinθ + b cosθ and y= acosθ - bsinθ, prove that x² + y² = a² + b²

Solution :-

1) x = asinθ + bcosθ

Squaring on both sides

⇒ x² = (asinθ + bsinθ)²

⇒ x² = a²sin²θ + b²cos²θ + 2ab.sinθ.cosθ ---eq(1)

2) y = acosθ - bsinθ

Squaring on both sides

⇒ y² = (acosθ - bsinθ)²

⇒ y² = a²cos²θ + b²sin²θ - 2ab.sinθ.cosθ --eq(2)

Adding eq(1) and eq(2)

⇒ x² + y² = a²sin²θ + b²cos²θ + 2ab.sinθ.cosθ + a²cos²θ + b²sin²θ - 2ab. sinθcosθ

⇒ x² + y² = a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ)

⇒ x² + y² = a²(1) + b²(1)

[ Because sin²θ + cos²θ = 1 ]

⇒ x² + y² = a² + b²

Hence proved.

Answer 2

Answer:

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