Question

plzzz ans....PROVE THE FOLLOWING

Answer 1

HELLO DEAR,


we know that:-

(a³-b³)=(a-b)(a²+b²+ab)
(a³+b³)=(a+b)(a²+b²-ab)
$\frac{ {sin}^{3} \alpha + {cos}^{3} \alpha }{ \sin \alpha + \cos \alpha } + \frac{ {sin}^{3} \alpha - {cos}^{3} \alpha }{ \sin \alpha - \cos \alpha } \\ \\ = \frac{(sin \alpha + cos \alpha )( {sin }^{2} \alpha + {cos}^{2} \alpha - \sin \alpha \cos \alpha )}{( \sin \alpha + \cos \alpha )} + \frac{(sin \alpha - cos \alpha )( {sin }^{2} \alpha + {cos}^{2} \alpha + \sin \alpha \cos \alpha )}{( \sin \alpha - \cos \alpha )} \\ \\ = (1 - sin \alpha cos \alpha ) + (1 + sin \alpha cos \alpha ) \\ \\ = 1 + 1 = 2$
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

=> LHS = [ ( sin³A + cos³A ) / ( sin A + cos A ) ] + [ ( sin³A - cos³A ) ] / ( sin A - cos A ) ]

=> [ (sin A + cos A) ( sin²A+cos²A - sinA cosA ) / ( sin A + cos A ) ] + [ (sinA - cosA) (sin²A+cos²A+sinA cosA) ] / (sin A - cos A )


=> sin A + cos A in numerator and denominator gets cancelled out

=> sin A- cos A in numerator and denominator got cancelled out


=> sin²A+cos²A - sinA cos A + sin²A+cos²A+sinA cosA

=> 1 + 1 = 2. ( -sin A cos A & + sin A cos A get cancelled out )


since , LHS= RHS

hence , PROVED

hope this helps