Answer
$\frac{uv-\sqrt {1-u^2}\sqrt {1-v^2}}{v\sqrt {1-u^2}+u\sqrt {1-v^2}}$,
$|u|\le1$,
$|v|\le1$.
Work Step by Step
Step 1. Letting $sin^{-1}u=x$, we have $sin(x)=u$ and $tan(x)=\frac{u}{\sqrt {1-u^2}}$ with $|u|\le1$
Step 2. Letting $cos^{-1}v=y$, we have $cos(y)=v$ and $tan(y)=\frac{\sqrt {1-v^2}}{v}$ with $|v|\le1$
Step 3. We have $tan(x-y)=\frac{tan(x)-tan(y)}{1+tan(x)tan(y)}=\frac{\frac{u}{\sqrt {1-u^2}}-\frac{\sqrt {1-v^2}}{v}}{1+\frac{u}{\sqrt {1-u^2}}\frac{\sqrt {1-v^2}}{v}}=\frac{uv-\sqrt {1-u^2}\sqrt {1-v^2}}{v\sqrt {1-u^2}+u\sqrt {1-v^2}}$
