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Question

Value it :-
if xsin3θ+ycos3θ=sinθcosθ and xsinθ=ycosθ ,prove thatx2+y2=1.

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Solution

Given-
xsin3θ+ycos3θ=sinθcosθ &

xsinθ=ysinθ

(xsinθ)(sin2θ)+ycos3θ=sinθcosθ

ycosθ(sin2θ)+ycos3θ=sinθcosθ

ycosθ(sin2θ+cos2θ)=sinθcosθ

ycosθ=sinθcosθ

y=sinθ

similarly,
x=cosθ

so,

x2+y2=sin2θ+cos2θ=1

Hence Proved.

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