Q.$x=a(\theta-\sin\theta)$, $y=a(1-\cos\theta)$

Given $x=a(\theta-\sin\theta)$, $y=a(1-\cos\theta)$.

Step 1: $\dfrac{dx}{d\theta}=a(1-\cos\theta)$ and $\dfrac{dy}{d\theta}=a\sin\theta$, so $\dfrac{dy}{dx}=\dfrac{\sin\theta}{1-\cos\theta}$.

Step 2 — simplify using $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$ and $1-\cos\theta=2\sin^2(\theta/2)$: $\dfrac{dy}{dx}=\dfrac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)}=\cot(\theta/2)$.

Step 3 — differentiate this w.r.t. $\theta$: $\dfrac{d}{d\theta}\left[\cot(\theta/2)\right]=-\dfrac12\csc^2(\theta/2)$. …

Parametric differentiation: find $dy/dx$ in terms of $\theta$ using half-angle identities, then diff …

Forgetting to divide by $dx/d\theta$ a second time — $d^2y/dx^2=\dfrac{d}{d\theta}(dy/dx)\div\dfrac{dx}{d\theta} …