Question

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Answer 1

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Answer 2

$= > \frac{ax}{cos \alpha } + \frac{by}{sin \alpha } = {a}^{2} - {b}^{2} \: \: \: ...eq _{1} \\ \\ = > \frac{axsin \alpha }{ {cos}^{2} \alpha } - \frac{bycos \alpha }{ {sin}^{2} \alpha } = 0 \\ \\ = > \frac{ {sin}^{3} \alpha }{by} = \frac{ {cos}^{3} \alpha }{ax} \\ \\ = > {( \frac{ {sin }^{ 3 } \alpha }{by}) }^{ \frac{2}{3} } = {( \frac{ {cos}^{3} \alpha }{ax}) }^{ \frac{2}{3} }$
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we know that :-

$if = > \: \: \frac{a}{b} = \frac{c}{d } \\ \\ then \: \: \: = > \frac{a}{b} = \frac{c}{d} = \frac{a + c}{b + d}$
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Then

$= > \frac{ {sin}^{2} \alpha }{ {(by)}^{ \frac{2}{3} } } = \frac{ {cos}^{2} \alpha }{ {(ax)}^{ \frac{2}{3} } } = \frac{ {sin}^{2} \alpha + {cos}^{2} \alpha }{ {(by)}^{ \frac{2}{3} } + {(ax)}^{ \frac{2}{3} } } \\ \\ = > \frac{ {sin}^{2} \alpha }{ {(by)}^{ \frac{2}{3} } } + \frac{ {cos}^{2} \alpha }{ {(ax)}^{ \frac{2}{3} } } = \frac{1}{ {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } } \\ \\ = > sin \alpha = \frac{ {(by)}^{ \frac{1}{3} } }{ \sqrt{ {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } } } \\ \\ and \\ \\ = > cos \alpha = \frac{ {(ax)}^{ \frac{1}{3} } }{ \sqrt{ {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } } } \\ \\ putting \: \: the\: \: value \: \: in \: \: eq1 \\ \\ = > {(ax)}^{ \frac{2}{3} } \sqrt{ {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } } \: + {(by)}^{ \frac{2}{3} } \sqrt{ {(ax)}^{ \frac{2}{3} } + {(b)}^{ \frac{2}{3} } } = {a}^{2} - {b}^{2} \\ \\ = > ( \sqrt{ {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } } ).( {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } ) = {a}^{2} - {b}^{2} \\ \\ = > {(( {ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } )}^{ \frac{3}{2} } = {a}^{2} - {b}^{2} \\ \\ = > {(ax)}^{ \frac{2}{3} } + {(by)}^{ \frac{2}{3} } = {( {a}^{2} - {b}^{2} ) }^{ \frac{2}{3} }$
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