Answer
$tanx+tany=\frac{sin(x+y)}{cosxcosy}$
Work Step by Step
We need to prove the identity
$tanx+tany=\frac{sin(x+y)}{cosxcosy}$
Let us solve the right side of the given identity.
$\frac{sin(x+y)}{cosxcosy}=\frac{sinxcosy+cosxsiny}{cosxcosy}$
$\frac{sin(x+y)}{cosxcosy}=\frac{sinx}{cosx}+\frac{siny}{cosy}$
Hence, $tanx+tany=\frac{sin(x+y)}{cosxcosy}$
