If x = a cos θ + b sin θ and y = a sin θ - b cos θ, prove that x 2 + y 2 = a 2 + b 2 .

x 2 + y 2 = (a cos θ + b sin θ) 2 + (a sin θ - b cos θ) 2 ⇒ x 2 + y 2 = a 2 cos 2 θ + b 2 sin 2 θ + 2ab cos θ. sin θ + a 2 sin 2 θ + b 2 cos 2 θ - 2ab cos θ. sin θ ⇒ x 2 + y 2 = a 2 (sin 2 θ + cos 2 θ) + b 2 (sin 2 θ + cos 2 θ) ⇒ x 2 + y 2 = (a 2 + b 2 )(sin 2 θ + cos 2 θ) As, sin 2 θ + cos 2 θ = 1. ⇒ x 2 + y 2 = (a 2 + b 2 ). Hence proved that x 2 + y 2 = (a 2 + b 2 ).

Mathematics

If x = a cos θ + b sin θ and y = a sin θ - b cos θ, prove that x² + y² = a² + b².

Trigonometrical Ratios

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Answer

x² + y² = (a cos θ + b sin θ)² + (a sin θ - b cos θ)²

⇒ x² + y² = a² cos² θ + b² sin² θ + 2ab cos θ. sin θ + a² sin² θ + b² cos² θ - 2ab cos θ. sin θ

⇒ x² + y² = a²(sin² θ + cos² θ) + b²(sin² θ + cos² θ)

⇒ x² + y² = (a² + b²)(sin² θ + cos² θ)

As, sin² θ + cos² θ = 1.

⇒ x² + y² = (a² + b²).

Hence proved that x² + y² = (a² + b²).

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If x = a cos θ + b sin θ and y = a sin θ - b cos θ, prove that x² + y² = a² + b².

Trigonometrical Ratios

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