(sin theta+cos theta)^{2}+(sin theta-cos thet | vedclass

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(B) We are given the expression: $(\sin 	heta + \cos 	heta)^2 + (\sin 	heta - \cos 	heta)^2$.Using the algebraic identities $(a+b)^2 = a^2 + 2ab +

Vedclass · Accepted Answer

$2$

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(B) We are given the expression: $(\sin 	heta + \cos 	heta)^2 + (\sin 	heta - \cos 	heta)^2$.Using the algebraic identities $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$,we expand the terms:$= (\sin^2 	heta + 2 \sin 	heta \cos 	heta + \cos^2 	heta) + (\sin^2 	heta - 2 \sin 	heta \cos 	heta + \cos^2 	heta)$.Combining like terms,the $2 \sin 	heta \cos 	heta$ and $-2 \sin 	heta \cos 	heta$ cancel out:$= \sin^2 	heta + \cos^2 	heta + \sin^2 	heta + \cos^2 	heta$.$= 2 \sin^2 	heta + 2 \cos^2 	heta$.$= 2(\sin^2 	heta + \cos^2 	heta)$.Since the fundamental trigonometric identity is $\sin^2 	heta + \cos^2 	heta = 1$,we substitute this value:$= 2(1) = 2$.

$(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2} = \dots$

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