Question

if x= a sin teta y = b tan teta then prove that a²/x²-b²/y²=1​

Answer 1

Given :

• x = a sinθ
• y = b tanθ

To prove :

$\sf \: \dfrac{ {a}^{2} }{ {x}^{2} } - \dfrac{ {b}^{2} }{ {y}^{2} } \: = \: 1$

Identity used :

• tanθ = sinθ /cosθ
• sin²θ + cos²θ = 1

Solution :

x = a sinθ

$\sf \implies \: \dfrac{a}{x} \: = \: \dfrac{1}{ \sin( \theta) } \: \: \: \: equation \: 1$

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y = b tanθ

$\sf \implies \: \dfrac{b}{y} \: = \: \dfrac{1}{ \tan( \theta) } \: \: \: \: equation \: 2$

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Now we need to prove that

$\sf \: \dfrac{ {a}^{2} }{ {x}^{2} } - \dfrac{ {b}^{2} }{ {y}^{2} } \: = \: 1$

Taking RHS = 1

Now Taking , LHS

$\sf \implies \: LHS \:=\dfrac{ {a}^{2} }{ {x}^{2} } - \dfrac{ {b}^{2} }{ {y}^{2} } \:$

$\sf \implies \: LHS \:= \: \bigg(\dfrac{ {a}}{ {x}} \bigg)^{2} - \: \bigg(\dfrac{ {b}}{ {y} }\bigg)^{2} \:$

Put value of (a/x) and (b/y) from equation 1 and 2 respectively

$\sf \implies \: LHS \:= \: \bigg(\dfrac{ {1}}{ { \sin( \theta) }} \bigg)^{2} - \: \bigg(\dfrac{ {1}}{ { \tan( \theta) } }\bigg)^{2} \:$

$\sf \implies \: LHS \:= \: \dfrac{ {1}}{ { \sin^{2} ( \theta) }} \: - \: \bigg(\dfrac{ { \cos( \theta) }}{ { \sin( \theta) } }\bigg)^{2} \:$

$\sf \implies \: LHS \:= \: \dfrac{ {1}}{ { \sin^{2} ( \theta) }} \: - \: \dfrac{ { \cos^{2} ( \theta) }}{ { \sin^{2} ( \theta) } }\:$

$\sf \implies \: LHS \:= \: \: \dfrac{ { 1 - \cos^{2} ( \theta) }}{ { \sin^{2} ( \theta) } }\:$

$\sf \implies \: LHS \:= \: \: \dfrac{ { \sin^{2} ( \theta) }}{ { \sin^{2} ( \theta) } }\:$

$\sf \implies \: LHS \:= \:1$

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RHS = 1 = LHS

Hence proved