The parametric equations of a curve are $$ x = t + \cos(t)$$$$ y= \ln(1+\sin(t))$$
where $-\frac {1}{2}\pi < t <\frac {1}{2}\pi$
$i) $ Show that $\frac{dy}{dx} = \sec (t)$
I got that $\frac{dx}{dt} = 1 -\sin(t)$ and $\frac{dy}{dt} = \frac {\cos(t)} {1 + \sin(t)}$, I don't get the correct answer when I do $\frac {dy}{dx} = \frac {dx}{dt} * \frac {dt}{dy}$, where am I going wrong?
You introduced $\frac {dy}{dx} = \frac {dx}{dt} * \frac {dt}{dy}$ which is wrong. Well, we have $$\begin{align} \frac{dy}{dx}&=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ &=\frac{\frac{\cos t}{1+\sin t}}{1-\sin t}\\ &=\frac{\cos t}{1+\sin t}\cdot\frac{1}{1+\sin t}\\ &=\frac{\cos t}{1-\sin^2t}\\ &=\frac{\cos t}{\cos^2t}\\ &=\frac{1}{\cos t}\\ &=\sec t \end{align}$$