If x=2 \cos \theta-\cos 2 \theta and y=2 \sin \theta- \sin 2 \theta, then \frac{d y}{d x} is

Correct Answer is : cosθ−cos2θsin2θ−sinθ

\frac{\cos \theta+\cos 2 \theta}{ \sin \theta- \sin 2 \theta}

\frac{\cos \theta-\cos 2 \theta}{ \sin 2 \theta- \sin \theta}

\frac{\cos \theta-\cos 2 \theta}{ \sin \theta- \sin 2 \theta}

\frac{\cos 2 \theta-\cos \theta}{ \sin 2 \theta+ \sin \theta}

CBSE Class 12 Math 2022 Term I Solved Paper - Question 22

Question : 22 of 50

Marks: +1, -0

x=2 \cos θ-\cos 2 θ

and

y=2 sin \;θ- sin \;2 θ

, then

\;{d y}/{d x}

Solution:

\;\text " Explanation: Given, "\;\;x=2 \cos θ-\cos 2 θ

\;\text " and "\;\;y\;=2 sin \;θ- sin \;2 θ

\;\;\text " Therefore, "\;\;\;{d x}/{d θ}\;=-2 sin \;θ+2 sin \;2 θ

\;\text " and "\;\;\;{d y}/{d θ}\;=2 \cos θ-2 \cos 2 θ

∴ \;\;{d y}/{d x}\;=\;{2 \cos θ-2 \cos 2 θ}/{-2 sin \;θ+2 sin \;2 θ}

\;\;\text " or "\;\;\;{d y}/{d x}\;=\;{\cos θ-\cos 2 θ}/{ sin \;2 θ- sin \;θ}

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