Answer
$\frac{u\sqrt {1-v^2}-v}{\sqrt {1+u^2}}$, $u\in(-\infty,\infty), |v|\le1$.
Work Step by Step
1. Let $tan^{-1}(u)=x$, we have $tan(x)=u, sin(x)=\frac{u}{\sqrt {1+u^2}}, cos(x)=\frac{1}{\sqrt {1+u^2}}$
2. Let $sin^{-1}(v)=y$, we have $sin(y)=v, cos(y)=\sqrt {1-v^2}$
3. Thus $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)=\frac{u\sqrt {1-v^2}}{\sqrt {1+u^2}}-\frac{v}{\sqrt {1+u^2}}=\frac{u\sqrt {1-v^2}-v}{\sqrt {1+u^2}}$ where $u\in(-\infty,\infty), |v|\le1$.
