Answer
$0$
Work Step by Step
I know that $\tan$ has a period of $180^\circ$, hence $\tan{\theta}=\tan{(\theta-180^\circ)}.$ Thus $\tan{200^\circ}=\tan{200^\circ-180^\circ}=\tan{20^\circ}$.
I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta-360^\circ)}.$ Thus $\cos{380^\circ}=\cos{(380^\circ-360^\circ)}=\cos{20^\circ}.$
I know that $\sin$ is an even function, which means $f(-\theta)=-f(\theta).$ Therefore $\sin{-20^\circ}=-\sin{20^\circ}$.
Thus: $\frac{\sin{-20^\circ}}{\cos{380^\circ}}+\tan{200^\circ}=\frac{-\sin{20^\circ}}{\cos{20^\circ}}+\tan{20^\circ}=-\tan{20^\circ}+\tan{20^\circ}=0$
Because $\frac{\sin{theta}}{\cos{theta}}=\tan{theta}$.
