## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

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Recall: Sine is an odd function so $\sin{(-\theta)} = - \sin{\theta}$. This means that $\sin{(-20^{\circ})} = - \sin{20^{\circ}}$ Since $\cos{(\theta-360^{\circ})} = \cos{\theta}$, then $\cos{380^{\circ}} = \cos{(380^{\circ}-360^{\circ})} = \cos{20^{\circ}}$ Thus, $\dfrac{\sin{(-20^{\circ})}}{\cos{380^{\circ}}} = \dfrac{- \sin{20^{\circ}}}{\cos{20^{\circ}}}$ Recall that $\tan{\theta} =\dfrac{\sin{\theta}}{\cos{\theta}}$. Hence, $\dfrac{- \sin{20^{\circ}}}{\cos{20^{\circ}}} = - \tan{20^{\circ}}$ Therefore, $\dfrac{\sin{(-20^{\circ})}}{\cos{380^{\circ}}} + \tan{20^{\circ}} = - \tan{20^{\circ}}+ \tan{20^{\circ}} =\boxed{0}$