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Question

If x=2 cos θ-cos 2 θ and y=2 sin θ-sin 2 θ, prove that dydx=tan 3 θ2

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Solution

We have, x=2 cosθ- cos2θ

dxdθ=2-sinθ--sin2θddθ2θdxdθ=-2sinθ+2 sin2θdxdθ=2sin2θ-sinθ ...iand,y=2 sinθ-sin2θ

dydθ=2 cosθ-cos2θddθ2θdydθ=2 cosθ-cos2θ2dydθ=2 cosθ-2 cos2θdydθ=2cosθ-cos2θ ...iiDividing equation ii by equation i,dydθdxdθ=2cosθ-cos2θ2sin2θ-sinθdydx=cosθ-cos2θsin2θ-sinθdydx=-2sinθ+2θ2sinθ-2θ22cos2θ+θ2sin2θ-θ2 sinA-sinB=2 cosA+B2sinA-B2and cosA-cosB=-2sinA+B2sinA-B2dydx=-sin3θ2sin-θ2cos3θ2sinθ2dydx=-sin3θ2-sinθ2cos3θ2sinθ2dydx=sin3θ2cos3θ2dydx=tan3θ2

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