Here, sinθ + cosθ = 1 Squaring on both sides, we get (sinθ + cosθ)² = 1 sin²θ + cos²θ + 2sinθ × cosθ = 1 1 + 2sinθ cosθ = 1 2sinθ× cosθ = 0 And we know that, sin2θ = 2sinθ cosθ ⇒ sin2θ = 0
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