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Q.

Ifxsin3θ+ycos3θ=cosθsinθandxsinθ=ycosθthenbyelimnatingθ

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a

x2+y2=2

b

x2+y2=4

c

x2+y2=3

d

x2+y2=1

answer is D.

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Detailed Solution

xsin3θ+ycos3θ=cosθsinθxycosθx3+ycos3θ=cosθycosθxy3+x2yx2cos3θ=ycos2θxcosθ=xx2+y2.Similarlysinθ=yx2+y2.  cos2θ+sin2θ=1x2(x2+y2)+y2(x2+y2)=1x2+y2=1.

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