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