Answer
See below.
Work Step by Step
Step 1. Letting $sin^{-1}v=x$, we have $sin(x)=v$ where $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$
Step 2. Letting $cos^{-1}v=y$, we have $cos(y)=v$ where $y\in[0,\pi]$
Step 3. We have $sin(x)=cos(y)=sin(\frac{\pi}{2}-y)$, thus $x=2k\pi+\frac{\pi}{2}-y$ and $x+y=2k\pi+\frac{\pi}{2}$
Step 4. As $(x+y)\in [-\frac{\pi}{2},\frac{3\pi}{2}]$, we have $k=0$ and $x+y=\frac{\pi}{2}$
