Question

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





plssssss anwer​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Trigonometric equations are

$\rm :\longmapsto\:\dfrac{xsin\theta }{a} - \dfrac{ycos\theta }{b} = 1 - - - (1)$

and

$\rm :\longmapsto\:\dfrac{xcos\theta }{a} + \dfrac{ysin\theta }{b} = 1 - - - (2)$

Now, on squaring equation (1), we get

$\rm :\longmapsto\: {\bigg[\dfrac{xsin\theta }{a} - \dfrac{ycos\theta }{b} \bigg]}^{2} = 1$

$\rm :\longmapsto\:\dfrac{ {x}^{2} {sin}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {cos}^{2}\theta }{ {b}^{2} } - 2\dfrac{xsin\theta cos\theta }{ab} = 1 - - - (3)$

Now, On squaring equation (2), we get

$\rm :\longmapsto\: {\bigg[\dfrac{xcos\theta }{a} + \dfrac{ysin\theta }{b} \bigg]}^{2} = 1$

$\rm :\longmapsto\:\dfrac{ {x}^{2} {cos}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {sin}^{2}\theta }{ {b}^{2} } + 2\dfrac{xsin\theta cos\theta }{ab} = 1 - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} {sin}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {cos}^{2}\theta }{ {b}^{2} } + \dfrac{ {x}^{2} {cos}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {sin}^{2}\theta }{ {b}^{2} } = 2$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} }( {cos}^{2}\theta + {sin}^{2}\theta ) + \dfrac{ {y}^{2} }{ {b}^{2} }( {cos}^{2}\theta + {sin}^{2}\theta ) = 2$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} }+ \dfrac{ {y}^{2} }{ {b}^{2} } = 2$

Hence, Proved

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given Trigonometric equations are

$\rm :\longmapsto\:\dfrac{xsin\theta }{a} - \dfrac{ycos\theta }{b} = 1 - - - (1)$

and

$\rm :\longmapsto\:\dfrac{xcos\theta }{a} + \dfrac{ysin\theta }{b} = 1 - - - (2)$

Now, on squaring equation (1), we get

$\rm :\longmapsto\: {\bigg[\dfrac{xsin\theta }{a} - \dfrac{ycos\theta }{b} \bigg]}^{2} = 1$

$\rm :\longmapsto\:\dfrac{ {x}^{2} {sin}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {cos}^{2}\theta }{ {b}^{2} } - 2\dfrac{xsin\theta cos\theta }{ab} = 1 - - - (3)$

Now, On squaring equation (2), we get

$\rm :\longmapsto\: {\bigg[\dfrac{xcos\theta }{a} + \dfrac{ysin\theta }{b} \bigg]}^{2} = 1$

$\rm :\longmapsto\:\dfrac{ {x}^{2} {cos}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {sin}^{2}\theta }{ {b}^{2} } + 2\dfrac{xsin\theta cos\theta }{ab} = 1 - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} {sin}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {cos}^{2}\theta }{ {b}^{2} } + \dfrac{ {x}^{2} {cos}^{2}\theta }{ {a}^{2} } + \dfrac{ {y}^{2} {sin}^{2}\theta }{ {b}^{2} } = 2$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} }( {cos}^{2}\theta + {sin}^{2}\theta ) + \dfrac{ {y}^{2} }{ {b}^{2} }( {cos}^{2}\theta + {sin}^{2}\theta ) = 2$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} }+ \dfrac{ {y}^{2} }{ {b}^{2} } = 2$

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1