Question

(1+sin theta)^2 + (1-cos theta)^2 = 3+2 (sin theta - cos theta)​

Answer 1

$LHS \\ = {(1 + \sin 0 ) }^{2} + {(1 - \cos0 )}^{2} \\ = 1 + {sin}^{2} 0 + 2sin0 + 1 + {cos}^{2}0 - 2cos0 \\ =1 + 1 + {sin}^{2}0 + {cos}^{2} 0 + 2sin0 - 2cos0 \\ = 2 + 1 + 2sin0 - 2cos0\\ = 3 + 2(sin0 - cos0)\\ = RHS$

Hence proved

Answer 2

Step-by-step explanation:

$\boxed{ {sin}^{2} \theta +{cos}^{2} \theta =1}\\\\LHS=(1+sin \theta)^2 + (1-cos \theta)^2 \\= 1 +2 sin\theta + {sin}^{2} \theta + 1 - 2cos \theta + {cos}^{2} \theta \\ = 2 + {sin}^{2} \theta +{cos}^{2} \theta + 2 sin\theta - 2cos \theta \\ =2 + 1 + 2 (sin\theta + cos \theta ) \\ = 3 + 2(sin\theta + cos \theta )\\=RHS\\ \\ hence \: it \: is \: proved$