If x sin3|+ y cos3|=sin|cos| and xsin|=ycos|, prove x2+y2=1. -Maths 9th

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Answer :

xsin3θ+ycos3θ=sinθcosθ (xsinθ)sin2θ+(ycosθ)cos2θ=sinθcosθ (xsinθ)sin2θ+(xsinθ)cos2θ=sinθcosθ xsinθ=sinθcosθ x=cosθ Again, ycosθ=xsinθ ycosθ=cosθsinθ y=sinθ Therefore, x2+y2=sin2θ+cos2θ=1.

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