Answer
See proof below.
Work Step by Step
1. Based on the figure from the problem, since triangles CDO and AOB are congruent, we have
$\bar {OC}=\bar {AB}=y$ and being in quadrant II gives the horizontal coordinate of D as $-y$
2. Similarly, $\bar {CD}=\bar {OA}=x$ and being in quadrant II gives the vertical coordinate of D as $x$
3. Since D $(-y,x)$ is the terminal point of ($t+\frac{\pi}{2})$, we have $sin(t+\frac{\pi}{2})=x=cos(t)$,
$cos(t+\frac{\pi}{2})=-y=-sin(t)$, and $tan(t+\frac{\pi}{2})=\frac{x}{-y}=-\frac{x}{y}=-cot(t)$,
