Question

Xcosa/a+ysina/b=1 and ax/(cosa)-by/(sina)=a²-b² then prove that x²/a²+y²/b²

Answer 1

||✪✪ CORRECT QUESTION ✪✪||

Xcosa/a+ysina/b=1 and xsinA/a - ycosA/b = 1,prove that x^2/a^2 + y^2/b^2=2 ?

|| ✰✰ ANSWER ✰✰ ||

→ Xcosa/a+ysina/b = 1 --------- Equation (1)

→ xsinA/a - ycosA/b = 1 ------- Equation (2)

Squaring both sides of The Equation and than adding Them ,, using (a+b)² = a² + b² + 2ab in First Equation and using (a-b)² = a² + b² - 2ab in Second Equation ,,

→ [ x²cos²A/a² + y² sin²A/b² + 2 * xcos/a * ysinA/b ] + [ x²sin²A/a² + y²cos²A/b² - 2 * xsinA/a * ycosA/b ] = 1² + 1²

→ x²cos²A /a² + x²sin²A/b² + x²sin²A/a² +y²cos²A/b² = 2

Rearranging the terms Now,

→ x²/a²(sin²A+ cos²A) + y²/b²(sin²A+cos²A) = 2

Now, using sin²A + cos²A = 1 ,

→ x²/a² + y²/b² = 2

✪✪ Hence Proved ✪✪

Answer 2

$\huge \boxed{solution}$

Refers to the attachment above .

$\boxed {formula \: used}$

$\mathtt{( {x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\mathtt{( {x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}$

$\mathtt{ \sin( { \theta}^{2} ) + { \cos( \theta) }^{2} = 1}$

Hope it helps