## Trigonometry (11th Edition) Clone

$$\tan(\pi-x)=-\tan x$$
$$\tan(\pi-x)$$ Apply the identity of tangent of a difference here, we have $$=\frac{\tan\pi-\tan x}{1+\tan\pi\tan x}$$ We know that $\tan\pi=0$. If you do not already know, see that $\tan\pi=\frac{\sin\pi}{\cos\pi}$, in which $\sin\pi$ equals $0$ and $\cos\pi$ equals $1$. That means, $$\tan(\pi-x)=\frac{0-\tan x}{1-0\times \tan x}$$ $$=\frac{-\tan x}{1}$$ $$=-\tan x$$