Answer
$\frac{1-uv}{\sqrt {1+u^2}\sqrt {1+v^2}}, u\in(-\infty,\infty),v\in(-\infty,\infty)$.
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 $tan^{-1}(v)=y$, we have $tan(y)=v, sin(y)=\frac{v}{\sqrt {1+v^2}}, cos(y)=\frac{1}{\sqrt {1+v^2}}$
3. Thus $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)=\frac{1}{\sqrt {1+u^2}}\cdot\frac{1}{\sqrt {1+v^2}}-\frac{u}{\sqrt {1+u^2}}\cdot\frac{v}{\sqrt {1+v^2}}=\frac{1-uv}{\sqrt {1+u^2}\sqrt {1+v^2}}, u\in(-\infty,\infty),v\in(-\infty,\infty)$.
