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please solve this question
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Hey mate!
We know that the square of the distance between two points is (x2-x1)^2 + (y2-y1)^2. So,
D = √[0-(sinθ-cosθ)]^2 + (sinθ+cos-0)^2
=> √sin^2θ + cos^2θ - 2sinθcosθ + sin^2θ + cos^2θ + 2sinθcosθ
=> √1+1 [Since sin^2θ + cos^2θ = 1]
=> √2.
That's the answer!
Hope it helps :)
We know that the square of the distance between two points is (x2-x1)^2 + (y2-y1)^2. So,
D = √[0-(sinθ-cosθ)]^2 + (sinθ+cos-0)^2
=> √sin^2θ + cos^2θ - 2sinθcosθ + sin^2θ + cos^2θ + 2sinθcosθ
=> √1+1 [Since sin^2θ + cos^2θ = 1]
=> √2.
That's the answer!
Hope it helps :)
Lekahdek:
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