Question

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Answer 1

Answer:

Step-by-step expl

Answer 2

Squaring both the equations and then adding ,

x²/a² cos²θ + y²/b² sin²θ + 2xy/ab sinθ.cosθ + x²/a² sin²θ + y²/b² cos²θ - 2xy/ab sinθ.cosθ = 2

⇒x²/a² (cos²θ + sin²θ) + y²/b² (sin²θ + cos²θ ) = 2

⇒x²/a² × 1 + y²/b² × 1 = 2 [ ∵ sin²x + cos²x = 1 from trigonometric identities ]

∴ x²/a² + y²/b² = 2 , hence proved

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