If x=cos theta and y=sin 5 theta,then left(1- | vedclass

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(D) Given,$x=\cos 	heta$ and $y=\sin 5 	heta$.Differentiating with respect to $	heta$:$\frac{d x}{d 	heta}=-\sin 	heta$ and $\frac{d y}{d 	heta}

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$-25$

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(D) Given,$x=\cos 	heta$ and $y=\sin 5 	heta$.Differentiating with respect to $	heta$:$\frac{d x}{d 	heta}=-\sin 	heta$ and $\frac{d y}{d 	heta}=5 \cos 5 	heta$.Therefore,$\frac{d y}{d x}=\frac{d y / d 	heta}{d x / d 	heta} = \frac{5 \cos 5 	heta}{-\sin 	heta} = -5 \cos 5 	heta \csc 	heta$.Now,find $\frac{d^2 y}{d x^2} = \frac{d}{d 	heta} \left( \frac{d y}{d x} ight) \cdot \frac{d 	heta}{d x}$.$\frac{d}{d 	heta} (-5 \cos 5 	heta \csc 	heta) = -5 [(-5 \sin 5 	heta) \csc 	heta + \cos 5 	heta (-\csc 	heta \cot 	heta)] = 25 \sin 5 	heta \csc 	heta + 5 \cos 5 	heta \csc 	heta \cot 	heta$.Multiplying by $\frac{d 	heta}{d x} = \frac{1}{-\sin 	heta} = -\csc 	heta$:$\frac{d^2 y}{d x^2} = -25 \sin 5 	heta \csc^2 	heta - 5 \cos 5 	heta \csc^2 	heta \cot 	heta$.Substitute into the expression $\left(1-x^2ight) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$:Since $1-x^2 = \sin^2 	heta$ and $x = \cos 	heta$:$\sin^2 	heta (-25 \sin 5 	heta \csc^2 	heta - 5 \cos 5 	heta \csc^2 	heta \cot 	heta) - \cos 	heta (-5 \cos 5 	heta \csc 	heta)$$= -25 \sin 5 	heta - 5 \cos 5 	heta \cot 	heta + 5 \cos 5 	heta \cot 	heta$$= -25 \sin 5 	heta = -25 y$.

If $x=\cos \theta$ and $y=\sin 5 \theta$,then $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)

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