Answer
See below.
Work Step by Step
1. Let $sin^{-1}v=x$, we have $sin(x)=v, cos(x)=\sqrt {1-v^2}$
2. Let $cos^{-1}v=y$, we have $cos(y)=v, sin(y)=\sqrt {1-v^2}$
3. $sin(x+y)=sin(x)cos(y)+cos(x)sin(y)=v(v)+\sqrt {1-v^2}\sqrt {1-v^2}=1$
4. Thus $x+y=2k\pi+\frac{\pi}{2}$
5. Consider the ranges of $x,y$, we have $x+y=\frac{\pi}{2}$
