Given, $x=a(t-\sin t)$ and $y=a\left(1+\cos t\right)$

$\therefore \frac{dx}{dt}=a-a\cos t \text{and }\frac{dy}{dt}=-a\sin t$

$\because \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{-\sin t}{1-\cos t}$

$\therefore \frac{dy}{dx}=\frac{-2\sin {\displaystyle \frac{t}{2}}\cos \frac{t}{2}}{2{\sin }^2\frac{t}{2}}=-\cot \frac{t}{2}$

Now,

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\times \frac{dt}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-1}{2}\left(-{\mathrm{cosec}}^2\frac{t}{2}\right)\times \frac{1}{a\left(1-\cos t\right)}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{1}{2}{\mathrm{cosec}}^2\frac{t}{2}\times \frac{1}{a\times 2{\sin }^2\frac{t}{2}}=\frac{1}{4a{\sin }^4\frac{t}{2}}$

Hence, option $\left(1\right)$ is correct.