Question

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Answer 1

GIVEN :–

$\\ \bf \longrightarrow x { \sin }^{3} ( \alpha ) + y { \sin}^{3}( \alpha ) = \sin( \alpha ) \cos( \alpha )$

$\\ \bf \longrightarrow x { \sin }( \alpha ) - y { \cos}( \alpha ) = 0$

TO FIND :–

• Value of x² + y² = ?

SOLUTION :–

• According to the question –

$\\ \bf \implies x { \sin }( \alpha ) - y { \cos}( \alpha ) = 0$

$\\ \bf \implies x { \sin }( \alpha ) = y { \cos}( \alpha )$

$\\ \bf \implies x = y \dfrac{ { \cos}( \alpha )}{{ \sin }( \alpha )} \:\:\:\:\:----eq.(1)$

• Put the value of 'x' in other given equation –

$\\ \bf \implies y \dfrac{ { \cos}( \alpha )}{{ \sin }( \alpha )} { \sin }^{3} ( \alpha ) + y { \sin}^{3}( \alpha ) = \sin( \alpha ) \cos( \alpha )$

$\\ \bf \implies y { \cos}( \alpha ){ \sin }^{2} ( \alpha ) + y { \sin}^{3}( \alpha ) = \sin( \alpha ) \cos( \alpha )$

$\\ \bf \implies y { \cos}( \alpha )\sin ( \alpha ) + y { \sin}^{2}( \alpha ) =\cos( \alpha )$

$\\ \bf \implies y\sin ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \} =\cos( \alpha )$

$\\ \bf \implies y = \dfrac{\cos( \alpha )}{\sin ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}}$

$\\ \large\implies{ \boxed{ \bf y = \dfrac{\cos( \alpha )}{\sin ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}}}}$

• Put the value of 'y' in eq.(1) –

$\\ \bf \implies x = \bigg \{\dfrac{\cos( \alpha )}{\sin ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}}\bigg \} \dfrac{ { \cos}( \alpha )}{{ \sin }( \alpha )}$

$\\ \large\implies{ \boxed{ \bf x = \dfrac{\cos^{2} ( \alpha )}{\sin^{2} ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}}}}$

• Now Let's find –

$\\ \implies \bf {x}^{2} + {y}^{2} = \bigg \{{\dfrac{\cos^{2} ( \alpha )}{\sin^{2} ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}} \bigg \}}^{2} + \bigg \{ {\dfrac{\cos( \alpha )}{\sin ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}}} \bigg \}^{2}$

$\\ \implies \bf {x}^{2} + {y}^{2} = {\dfrac{\cos^{2} ( \alpha )}{\sin^{2} ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}^{2} }} \bigg \{ \dfrac{\cos^{2} ( \alpha )}{\sin^{2} ( \alpha )}+1 \bigg \}$

$\\ \implies \bf {x}^{2} + {y}^{2} = {\dfrac{\cos^{2} ( \alpha )}{\sin^{2} ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}^{2} }} \bigg \{ \dfrac{\cos^{2} ( \alpha ) +\sin^{2} ( \alpha )}{\sin^{2} ( \alpha )} \bigg \}$

$\\ \implies \bf {x}^{2} + {y}^{2} = {\dfrac{\cos^{2} ( \alpha )}{\sin^{2} ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}^{2} }} \bigg \{ \dfrac{1}{\sin^{2} ( \alpha )} \bigg \}$

$\\ \implies \bf {x}^{2} + {y}^{2} = {\dfrac{\cos^{2} ( \alpha )}{\sin^{4} ( \alpha )\{{ \cos}( \alpha ) + \sin ( \alpha ) \}^{2} }}$

$\\ \implies \bf {x}^{2} + {y}^{2} = {\dfrac{\cos^{2} ( \alpha )}{\sin^{4} ( \alpha )\{{ \cos} ^{2} ( \alpha ) + \sin ^{2} ( \alpha ) + 2 \cos( \alpha ) \sin( \alpha ) \}}}$

$\\ \implies \bf {x}^{2} + {y}^{2} = {\dfrac{\cot^{2} ( \alpha )}{\sin^{2} ( \alpha )\{1+ 2 \cos( \alpha ) \sin( \alpha ) \}}}$

$\\ \implies \large{ \boxed{\bf {x}^{2} + {y}^{2} = {\dfrac{\cot^{2} ( \alpha ) cosec^{2} ( \alpha ) }{1+ 2 \cos( \alpha ) \sin( \alpha ) }}}}$

Answer 2

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Refer to attachment.

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