Question

$\sin( \alpha ) + \cos( \alpha ) = \sqrt{2} \cos( \alpha )$
then find
$\cos( \alpha ) - \sin( \alpha ) =$
​

Answer 1

$\huge\green{\mathbb{\underline{\underline{ANSWER}}}}$

GIVEN,

$\sin( \alpha ) + \cos( \alpha ) = \sqrt{2} \cos( \alpha )$

-------------(1)

${\mathrm{\underline{\underline\red{We \: Need \: To \: Find \: That \: :-}}}}$

$let \\ \cos( \alpha ) - \sin( \alpha ) = x$

--------------(2)

taking equation (1) and S.O.B.S(squaring on both sides)

${( \sin( \alpha ) + \cos( \alpha )) }^{2} = {( \sqrt{2} \cos( \alpha ) ) }^{2} \\ { \sin( \alpha ) }^{2} + 2 \sin( \alpha ) .\cos( \alpha ) + { \cos( \alpha ) }^{2} = 2 { \cos( \alpha ) }^{2}$

assuming this equation as equation (3)

taking equation (2) and S.O.B.S(squaring on both sides)

${( \cos( \alpha ) - \sin( \alpha ) )}^{2} = {x}^{2} \\ { \cos( \alpha ) }^{2} - 2 \sin( \alpha) . \cos( \alpha ) + { \sin( \alpha ) }^{2} = {x}^{2}$

assuming this equation as equation (4)

now,

adding equation (3)+equation (4)

then,

$\: \: \: \: \: { \sin( \alpha ) }^{2} + 2 \sin( \alpha ) . \cos( \alpha ) + { \cos( \alpha ) }^{2} = 2 { \cos( \alpha ) }^{2} \\ + {\underline{\underline{ { \sin( \alpha ) }^{2} - 2 \sin( \alpha ). \cos( \alpha ) + { \cos( \alpha ) }^{2} = {x}^{2} }}} \\ </p><p></p><p></p><p> \: \: \: \: 2 { \sin( \alpha ) }^{2} + 2 { \cos( \alpha ) }^{2} = 2 { \cos( \alpha ) }^{2} + {x}^{2} \\ \\ \: \: \: \: 2 { \sin( \alpha ) }^{2} + 2 { \cos( \alpha ) }^{2} - 2 { \cos( \alpha ) }^{2} = {x}^{2} \\ \: \: \: \: {x}^{2} = 2 { \sin( \alpha ) }^{2} \\ \: \: \: \: x = \sqrt{2 { \sin( \alpha ) }^{2} } \\ \: \: \: \: x = \sqrt{2} . \sqrt{ { \sin( \alpha ) }^{2} }$

${\underline{\underline\red{\mathrm{x = \sqrt{2} \sin( \alpha ) }}}}$

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