Math, asked by suzanneshaikh14, 2 months ago

If x/a cos +y/b sin =1 and x/a sin - y/b cos =1,prove that x2/a2+y2/b2=2​

Answers

Answered by amansharma264
10

EXPLANATION.

⇒ x/a cosθ + y/b sinθ = 1. - - - - - (1).

⇒ x/a sinθ - y/b cosθ = 1. - - - - - (2).

As we know that,

Square the equation (1), we get.

⇒ [x/a cosθ + y/b sinθ]² = [1]².

As we know that,

Formula of :

⇒ (x + y)² = x² + y² + 2xy.

Using this formula in equation, we get.

⇒ [x²/a² cos²θ + y²/b² sin²θ + 2xy/ab sinθcosθ] = 1. - - - - - (3).

Squaring the equation (2), we get.

⇒ [x/a sinθ - y/b cosθ]² = [1]².

As we know that,

Formula of :

⇒ (x - y)² = x² + y² - 2xy.

Using this formula in equation, we get.

⇒ [x²/a² sin²θ + y²/b² cos²θ - 2xy/ab sinθcosθ] = 1. - - - - - (4).

From equation (3) & (4) adding, we get.

⇒ [x²/a² cos²θ + y²/b²sin²θ + 2xy/ab sinθcosθ + x²/a²sin²θ + y²/b² cos²θ - 2xy/ab sinθcosθ] = 1 + 1.

⇒ [x²/a² cos²θ + x²/a² sin²θ + y²/b² sin²θ + y²/b² cos²θ] = 1 + 1.

⇒ [x²/a²(cos²θ + sin²θ) + y²/b²(sin²θ + cos²θ)] = 1 + 1.

As we know that,

Formula of :

⇒ sin²θ + cos²θ = 1.

Using this formula in equation, we get.

⇒ [x²/a² + y²/b²] = 2.

Hence proved.

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