Question

sin theta - cos theta = root 2 sin theta then prove that sin theta + cos theta = root 2 cos theta

Answer 1

Solution...


$\cos( \alpha ) - \sin( \alpha ) = \sqrt{2} \sin( \alpha ) .................. given$


$\cos( \alpha ) = \sqrt{2} \sin( \alpha ) + \sin( \alpha ) \\ \\ \cos( \alpha ) = ( \sqrt{2} + 1) \sin( \alpha ) \\ \\ \frac{1}{ \sqrt{2} + 1 } = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$



Now Rationalise LHS


$\frac{1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1} = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \\ \frac{ \sqrt{2} - 1}{ { (\sqrt{2} \: )}^{2} - {1}^{2} } = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$


$\frac{ \sqrt{2 } - 1}{2 - 1} = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \\ (\sqrt{2 } - 1) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \\ \cos( \alpha ) ( \sqrt{2} - 1) = \sin( \alpha )$


$\sqrt{2} \cos( \alpha ) - \cos( \alpha ) = \sin( \alpha ) \\ \\ \sqrt{2} \cos( \alpha ) = \sin( \alpha ) + \cos( \alpha ) \\ \\ \\ therefore............ \\ \\ \\ \sin( \alpha ) + \cos( \alpha ) = \sqrt{2} \cos( \alpha )$
$\\ \\ \\ hence \: proved......... \\ \\ \\ hope \: this \: will \: help \: you \: ...................... \\ \\ \\ \\ \\ \\ be \: brainly................$



☞ ⛧⛧ Ⓜr. Thakur ⛧⛧

Answer 2

hope this help you buddy
with lots of love❤