Math, asked by babitachavan023, 6 months ago

if x=asin theta y=bcos theta eliminate theta​

Answers

Answered by BrainlyIAS
9

x = a sinθ

Squaring on both sides , we get ,

⇒ x² = ( a sinθ )²

⇒ x² = a² sin²θ

x²/a² = sin²θ __ (1)

y = b cosθ

Squaring on both sides ,

⇒ y² = ( b cosθ )²

⇒ y² = b² cos²θ

y²/b² = cos²θ __ (2)

On adding (1) and (2) , we get ,

⇒ x²/a² + y²/b² = sin²θ + cos²θ

x²/a² + y²/b² = 1

[ From Trigonometric identity , sin²x + cos²x = 1 ]

Additionally , this equation x²/a² + y²/b² = 1 is called ellipse equation .


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MystícPhoeníx: Well Explained !! :)
Answered by Anonymous
27

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\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}

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