if the

It is given that, $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$

$\therefore \frac{d x}{d t}=a \cdot \frac{d}{d t}(\cos t+t \sin t)$

$=a\left[-\sin t+\sin t \cdot \frac{d}{d x}(t)+t \cdot \frac{d}{d t}(\sin t)\right]$\

$=a[-\sin t+\sin t+t \cos t]=a t \cos t$

$\frac{d y}{d t}=a \cdot \frac{d}{d t}(\sin t-t \cos t)$

$=a\left[\cos t-\left\{\cos t \cdot \frac{d}{d t}(t)+t \cdot \frac{d}{d t}(\cos t)\right\}\right]$

$=a[\cos t-\{\cos t-t \sin t\}]=a t \sin t$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a t \sin t}{a t \cos t}=\tan t$

Then, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tan t)=\sec ^{2} t \cdot \frac{d t}{d x}$

$=\sec ^{2} t \cdot \frac{1}{a t \cos t}$    $\left[\frac{d x}{d t}=a t \cos t \Rightarrow \frac{d t}{d x}=\frac{1}{a t \cos t}\right]$

$=\frac{\sec ^{3} t}{a t}, 0