If x = a(cos t + log tan t/2), y = a sin t, find d²y/dt² and d²y/dx²

x = a(cos t + log tan t/2), y = a sin t
dy/dt = a cos t
d²y/dt² = -a sin t
dx/dt = a\begin{pmatrix} -sin t + \frac{sec^2 t/2}{2 tan t/2} \end{pmatrix} = a \frac{-2sin t tan t/2 + sec^2 t/2}{2 tan t/2} = a \frac{-2 sin t sin t/2 + 1}{cos t/2 sin t/2} = a \frac{1 - 2 sin t sin t/2}{cos t/2 sin t/2}

\frac{d^2y}{dx^2} = \frac{d(\frac{dy}{dx})}{dx} = \frac{d(\frac{dy}{dt} \times \frac{dt}{dx})}{dt} \times \frac{dt}{dx}
\therefore \frac{d^2y}{dx^2} = \frac{d(a cos t \times \frac{sin t}{a cos^2 t})}{dt} \times \frac{sin t}{a cos^2 t} = \frac{d(tan t)}{dt} \times \frac{sin t}{a cos^2 t} = sec^2 t \times \frac{sin t}{a cos^2 t}
\frac{d^2y}{dx^2} = \frac{sin t}{a cos^4 t}