## Precalculus (6th Edition) Blitzer

Let us consider the left side: $\tan x+\tan y$ We know that, $\tan x=\frac{\sin x}{\cos x}$. It implies: \begin{align} & \tan x+\tan y=\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y} \\ & =\frac{\sin x\cos y+\sin y\cos x}{\cos x\cos y} \end{align} By using the identity $\sin \left( x+y \right)=\sin x\cos y+\sin y\cos x$, we obtain: $\tan x+\tan y=\frac{\sin \left( x+y \right)}{\cos x\cos y}$ Thus, $\tan x+\tan y=\frac{\sin \left( x+y \right)}{\cos x\cos y}$