How much work will you put into this problem?

Note that $\frac{\tan t}{1 + \sec t}\cdot \frac{\cos t}{\cos t} = \frac{\sin t}{\cos t + 1},$ showing that $x(t) = y(t)$ . Therefore, $dy/dx = 1$ wherever they exist, and $d^2y/dx^2 = 0$ .

However, when $t = \pi$ the function is not defined because both $x(\pi)$ and $y(\pi)$ become $0/0$ . Can this discontinuity can be resolved? Consider the limit $\lim_{t\to\pi} x(t) = \lim_{t\to\pi} \frac{\sin t}{1 + \cos t} = \lim_{t\to\pi} \frac{\cos t}{-\sin t} = -\infty.$ (Here I used L'Hopital)

Thus when $t\to\pi$ , the point $(x,y)$ runs away to infinity, and $d^2y/dx^2$ is not defined.