Answer
$$\tan(180^\circ+\theta)=\tan\theta$$
Work Step by Step
$$X=\tan(180^\circ+\theta)$$
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,
$$\tan(180^\circ+\theta)=\tan\theta$$
