Question

सिद्ध कीजिए कि (sin A – cos A)² + 2 sin A cos A = 1

Answer 1

Step-by-step explanation:

(SinA-CosA)² + 2SinA*CosA

(Sin²A + Cos²A - 2SinA*CosA) + 2SinA*CosA

Sin²A + Cos²A = 1

Answer 2

$\red{ \huge\mathfrak{solution}}$

To prove :

$\green{ \boxed {{(sin \theta - cos \theta)}^{2} + 2sin \theta \: cos \theta = 1}}$

LHS

$\implies \: ( {sin \theta - cos \theta)}^{2} + 2 \: sin \theta \: cos \theta\\ \\ \implies \: {sin}^{2} \theta \: + {cos}^{2} \theta \: - 2sin \theta \: cos \theta \: + 2 sin \theta cos \theta \\ \\ \implies \: 1 - 2sin \theta \:cos \theta \: + 2sin \theta \: cos \theta \\ \\ \implies 1 \cancel { - 2sin \theta \: cos \theta} \cancel{ + 2sin \theta cos \theta} \\ \\ \implies \: 1$

LHS = RHS

$\huge \boxed{ \purple{ \bold{proved}}}$

Formula used :

◆

$\huge \boxed{ \blue{ \bold{{sin}^{2} \theta + {cos}^{2} \theta = 1}}}$

◆

$\boxed{ \red{ \bold{ {(x - y)}^{2} \times {x}^{2} + {y}^{2} - 2xy}}}$