Answer
$\frac{\sqrt {1+u^2}}{v-u\sqrt {1-v^2}}, u\in(-\infty,\infty), |v|\le1$.
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 $cos^{-1}(v)=y$, we have $cos(y)=v, sin(y)=\sqrt {1-v^2}, $
3. Thus $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)=\frac{v}{\sqrt {1+u^2}}-\frac{u\sqrt {1-v^2}}{\sqrt {1+u^2}}=\frac{v-u\sqrt {1-v^2}}{\sqrt {1+u^2}}$
4 And $sec(x+y)=\frac{1}{cos(x+y)}=\frac{\sqrt {1+u^2}}{v-u\sqrt {1-v^2}}, u\in(-\infty,\infty), |v|\le1$.
