If x = sin theta and y = cos(p theta),then (1 | vedclass

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(A) Given,$x = \sin 	heta$ and $y = \cos(p 	heta)$.$\frac{dx}{d	heta} = \cos 	heta$ and $\frac{dy}{d	heta} = -p \sin(p 	heta)$.$\frac{dy}{dx} =

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$x y_1 - p^2 y$

Vedclass · Answer

(A) Given,$x = \sin 	heta$ and $y = \cos(p 	heta)$.$\frac{dx}{d	heta} = \cos 	heta$ and $\frac{dy}{d	heta} = -p \sin(p 	heta)$.$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-p \sin(p 	heta)}{\cos 	heta}$.Since $\sin(p 	heta) = \sqrt{1 - y^2}$ and $\cos 	heta = \sqrt{1 - x^2}$,we have $\frac{dy}{dx} = -p \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}$.$\sqrt{1 - x^2} y_1 = -p \sqrt{1 - y^2}$.Squaring both sides: $(1 - x^2) y_1^2 = p^2 (1 - y^2)$.Differentiating with respect to $x$:$(1 - x^2) \cdot 2 y_1 y_2 + y_1^2 (-2x) = p^2 (-2y y_1)$.Dividing by $2 y_1$ (assuming $y_1 
eq 0$):$(1 - x^2) y_2 - x y_1 = -p^2 y$.Therefore,$(1 - x^2) y_2 = x y_1 - p^2 y$.Thus,option $A$ is correct.

If $x = \sin \theta$ and $y = \cos(p \theta)$,then $(1 - x^2) y_2 =$

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