Question

(sinθ+cosθ)²+(sinθ-cosθ)²=​

Answer 1

Answer:

(sinθ+cosθ)²+(sinθ-cosθ)²=2

Step-by-step explanation:

=sin^2 θ + cos^2 θ +2sinθ cosθ +sin^2 θ + cos^2 θ -2sinθ cosθ

sin^2 θ + cos^2 θ=1

=1+1

(sinθ+cosθ)²+(sinθ-cosθ)²=2

Answer 2

$\huge\tt{Answer:-}$

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

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$\huge\tt{Solution:-}$

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

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$\sf By \: using \: identinty \: (a+b)^{2} = a^{2} + b^{2} + 2ab \: and \: (a-b)^{2} = a^{2} + b^{2} - 2ab$

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

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$\sf Now, \: we \: know \: that \: sin^{2} \theta + cos^{2} \theta = 1$

$\sf : \implies (1 + 2sin \theta cos \theta) + (1 - 2sin \theta cos \theta)$

$\sf : \implies 1 + 2sin \theta cos \theta + 1 - 2sin \theta cos \theta$

$\sf : \implies 1 + \cancel{2sin \theta cos \theta} + 1 \: \cancel{- 2sin \theta cos \theta }$

$\sf : \implies 1 + 1$

$\sf : \implies 2$

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