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Question

If xsin3θ+ycos3θ=sinθcosθ and xsinθ=ycosθ. Find the value of x2+y2.

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Solution

xsinθ=ycosθ
x=ycosθsinθ
xsin3θ+ycos3θ=sinθcosθ
ycosθ=sinθcosθ
y=sinθ
x=ycosθsinθ=cosθ
x2+y2=sin2θ+cos2θ=1

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