Question

plz answer me this question...

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\underline{\textsf{Given,}} \\ \\ \sf \implies \dfrac{x}{a} sin \: \theta \: - \: \dfrac{y}{b}cos \: \theta \: = \: 1 \\ \\ \textsf{Squaring both sides ,} \\ \\ \\ \sf \implies \left ( \dfrac{x}{a} sin \: \theta \: - \: \dfrac{y}{b}cos \: \theta \right)^2 \: = \: 1^2 \\ \\ \\ \sf \implies \left( \dfrac{x}{a} sin \: \theta \right)^{2} \: + \: \left( \dfrac{y}{b} cos \: \theta \right)^{2} \: - \: 2 \: \dfrac{x}{a} sin \: \theta \: \times \: \dfrac{y}{b} cos \: \theta \: = \: 1 \\ \\ \\ \sf \implies \dfrac{ {x}^{2} }{ {a}^{2} } {sin}^{2} \theta \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: \: - \: \dfrac{2xy}{ab} sin \: \theta \: cos \: \theta \: = \: 1 \qquad...(1)$




$\underline{\textsf{And,}} \\ \\ \sf \implies \dfrac{x}{a} cos \: \theta \: + \: \dfrac{y}{b}sin \: \theta \: = \: 1 \\ \\ \textsf{Squaring both sides ,} \\ \\ \\ \sf \implies \left ( \dfrac{x}{a} cos \: \theta \: + \: \dfrac{y}{b}sin \: \theta \right)^2 \: = \: 1^2 \\ \\ \\ \sf \implies \left( \dfrac{x}{a} cos \: \theta \right)^{2} \: + \: \left( \dfrac{y}{b} sin \: \theta \right)^{2} \: + \: 2 \: \dfrac{x}{a} cos \: \theta \: \times \: \dfrac{y}{b} sin\: \theta \: = \: 1 \\ \\ \\ \sf \implies \dfrac{ {x}^{2} }{ {a}^{2} } {cos}^{2} \theta \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } {sin}^{2} \theta\: \: + \: \dfrac{2xy}{ab} sin \: \theta \: cos \: \theta \: = \: 1 \qquad...(2)$





$\underline{\textsf{Add both the equations ,}} \\ \\ \\ \sf \implies \dfrac{ {x}^{2} }{ {a}^{2} } {sin}^{2} \theta \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: \: - \: \cancel{\: \dfrac{2xy}{ab} sin \: \theta \: cos \: \theta \: } \: + \: \dfrac{ {x}^{2} }{ {a}^{2} } {cos}^{2} \theta \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } {sin}^{2} \theta\: \: + \: \cancel{\: \dfrac{2xy}{ab} sin \: \theta \: cos \: \theta \: } \: = \: 2 \\ \\ \\ \sf \implies\dfrac{ {x}^{2} }{ {a}^{2} } {sin}^{2} \theta \: + \: \dfrac{ {x}^{2} }{ {a}^{2} } {cos}^{2} \theta \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } {sin}^{2} \: = \: 2 \\ \\ \\ \sf \implies \dfrac{ {x}^{2} }{ {a}^{2} } ( {sin}^{2} \theta \: + \: {cos}^{2} \theta) \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } ( {sin}^{2} \theta \: + \: {cos}^{2} \theta) \: = \: 2 \\ \\ \\ \sf \implies \dfrac{ {x}^{2} }{ {a}^{2} } \: \times \: 1 \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } \: \times \: 1 \: = \: 2 \\ \\ \\ \: \: \sf \therefore \: \: \dfrac{ {x}^{2} }{ {a}^{2} } \: + \: \dfrac{ {y}^{2} }{ {b}^{2} } \: = \: 2 \\ \\ \\ \\ \underline{\mathsf{\Large{Proved !! }}}$



$\underline{\textsf{Trigonometric Identity Used : }} \\ \\ \sf \implies sin^2 \theta \: + \: cos^2 \theta \: = \: 1 \\ \\ \underline{\textsf{Algebraic Identity Used : }} \\ \\ \sf \implies (a \: - \: b)^2 \: = \: a^2 \: + \: b^2 \: - \: 2ab \\ \\ \sf \implies (a \: + \: b)^2 \: = \: a^2 \: + \: b^2 \: + \: 2ab$

Answer 2

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