**Given,**

x = a (cos t + t sin t)

**Differentiating both sides with respect to t, we get**

\(\frac{dx}{dt}\) = a (-sin t + t cos t + sin t)

⇒ \(\frac{dx}{dt}\) = at cos t **...(i)**

**Differentiating again with respect to t, we get**

\(\frac{d^2x}{dt^2}\)** **=** **a (- t sin t + t cos t)

=** **a (cos t + t sin t)

**Again,**

y = a (sin t - cos t)

**Differentiating with respect to t, we get**

\(\frac{dy}{dt}=\)** **a (cos t - t sin t - cos t) ..**.(ii)**

⇒ \(\frac{dy}{dt}=\) at sin t

**Differentiating again with respect to t, we get**

\(\frac{d^2y}{dt^2}\)** **= ** **a (t cos t +sin t)

**Now,**

\(\frac{dy}{dx}=\) \(\frac{dt/dy}{dt/dy}\)

**[from (i) and (ii)]**

⇒ \(\frac{dy}{dx}=\)** \(\frac{at\,sin\,t}{at\,cos\,t}\) **

⇒ \(\frac{dy}{dx}=\) tan t

**Differentiating again with respect to x, we get**

\(\frac{d^2y}{dx^2}=\) sec^{2 }t.**\(\frac{dt}{dx}\) **