Answer :

Given: cos θ + sin θ = √2sin θ

Consider (sin θ + cos θ)2 + (sin θ – cos θ)2 = sin2 θ + cos2 θ + 2sin θ cos θ + sin2 θ + cos2 θ –

2 sin θ cos θ

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2sin2 θ + 2 cos2 θ

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2(sin2 θ + cos2 θ)

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2

(√2sin θ)2 + (sin θ – cos θ)2 = 2

(sin θ – cos θ)2 = 2 – 2sin2 θ

(sin θ – cos θ)2 = 2(1 – sin2 θ)

(sin θ – cos θ)2 = 2(cos2 θ)

(sin θ – cos θ) = ± √2 cos θ

Hence, proved.

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