If $x=2 \sin \theta$ and $y=2 \cos \theta$, then the value of $\frac{d^2 y}{d x^2}$ at $\theta=0$ is

$\frac{-1}{2}$

$x=2 \sin \theta ~~~y=2 \cos \theta$

so  $\frac{d x}{d \theta}=2 \cos \theta, \frac{d y}{d \theta}=-2\sin \theta$

$\Rightarrow \frac{d y / d \theta}{d x / d \theta}=\frac{-2 \sin \theta}{2 \cos \theta}=-\tan \theta$

so  $\frac{d^2 y}{d x^2}=-\sec ^2 \theta \frac{d \theta}{d x}$

$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-\sec ^2 \theta}{2 \cos \theta}$

$\Rightarrow\left.\frac{d^2 y}{d x^2}\right]_{\theta=0}=\frac{-1}{2}$

Option: 1