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

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Aicharya · Accepted Answer

Given, $x=\sin \theta$ and $y=\cos p \theta$ $\therefore \quad \frac{d y}{d \theta}=-p \sin p \theta$ and $\frac{d x}{d \theta}=\cos \theta$ $\Rightarrow \quad \frac{d y}{d x}=-\frac{p \sin p \theta}{\cos \theta}=-p \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$ $\Rightarrow \sqrt{1-x^2} \frac{d y}{d x}=-p \sqrt{1-y^2}$ $\Rightarrow \quad\left(\mathrm{l}-x^2\right)\left(\frac{d y}{d x}\right)^2=p^2\left(\mathrm{l}-y^2\right)$ (On squaring both sides) $\Rightarrow-2 x\left(\frac{d y}{d x}\right)^2+\left(\mathrm{l}-x^2\right) 2 \frac{d y}{d x}\left(\frac{d^2 y}{d x^2}\right)=-p^2 2 y \frac{d y}{d x}$ (On differentiating both sides) $\Rightarrow \quad-x \frac{d y}{d x}+\left(1-x^2\right) \frac{d^2 y}{d x^2}=-p^2 y$ $\Rightarrow \quad\left(\mathrm{l}-x^2\right) y_2=x y_1-p^2 y$ Hence, option (a) is correct.

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

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