## Trigonometry (11th Edition) Clone

$$\tan(180^\circ+\theta)=\tan\theta$$
$$X=\tan(180^\circ+\theta)$$ According to tangent sum identity: $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ Expand $X$: $$X=\frac{\tan180^\circ+\tan\theta}{1-\tan180^\circ\tan\theta}$$ At $180^\circ$, $\sin180^\circ=0$ and $\cos180^\circ=-1$. Therefore, $\tan180^\circ=\frac{\sin180^\circ}{\cos180^\circ}=0$ $$X=\frac{0+\tan\theta}{1-0\times\tan\theta}$$ $$X=\frac{\tan\theta}{1}$$ $$X=\tan\theta$$ Overall, $$\tan(180^\circ+\theta)=\tan\theta$$