Question

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Answer 1

Answer:

x^2 + y^2 = a^2 + b^2

Step-by-step explanation:

Here,

x = acosθ + bsinθ

    Square on both sides:

⇒ x² = ( acosθ + bsinθ )²

⇒ x² = a²cos²θ + b²sin²θ + 2ab.cosθ.sinθ

 

   Similarly,

y = asinθ - bcosθ

        Square on both sides:

⇒ y² = ( asinθ - bcosθ )²

⇒ y² = a²sin²θ + b²cos²θ - 2ab.cosθ.sinθ

  Adding x² and y²:

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

    = a²cos²θ + b²sin²θ + 2ab.cosθ.sinθ + a²sin²θ + b²cos²θ - 2ab.cosθ.sinθ

    = a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ

    = a²( cos²θ + sin²θ ) + b²( cos²θ + sin²θ )

    = a²( 1 ) + b²( 1 )

    = a² + b²

 Hence proved.

Answer 2

Question -

$\sf{ If, \ x = a \cos{ \theta } + b \sin{ \theta } \ and \ y = a \sin{ \theta } - b \cos{ \theta } } \\ \\ \sf{ \bold{ To \ Prove \ - x^2 + y^2 = a^2 + b^2 }} \\ \\ \sf{ We \ know \ the \ following \ identity \ - } \\ \\ \sf{ { \sin{ \theta } }^2 + { \cos{ \theta } }^2 = 1 } \\ \\ \sf{ \bold{ LHS \ - }} \\ \\ \sf{ x^2 = a^2 { \cos{ \theta } }^2 + b^2 { \sin{ \theta } }^2 + 2ab \cos{ \theta } \sin{ \theta } } \\ \\ \sf{ y^2 = a^2 { \sin{ \theta } }^2 + b^2 { \cos{ \theta } }^2 - 2ab \cos{ \theta } \sin{ \theta } } \\ \\ \sf{ x^2 + y^2 \ : }$

$\sf{ => a^2 { \cos{ \theta } }^2 + b^2 { \sin{ \theta } }^2 + a^2 { \sin{ \theta } }^2 + b^2 { \cos{ \theta } }^2 } \\ \\ \sf{ => a^2 ( { \sin{ \theta } }^2 + { \cos{ \theta } }^2 ) + b^2 ( { \sin{ \theta } }^2 + { \cos{ \theta } }^2 ) } \\ \\ \sf{ => a^2 + b^2 } \\ \\ \sf{\bold{ Hence \ Proved \ }}$