## Calculus: Early Transcendentals 8th Edition

$tanx+tany=\frac{sin(x+y)}{cosxcosy}$
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, $tanx+tany=\frac{sin(x+y)}{cosxcosy}$