Answer
$0$
Work Step by Step
Recall:
$\cos{\theta} = \cos{(\theta+360^{\circ})}$
This means that
$\cos{(-430^{\circ})} = \cos{(-430^{\circ}+360^{\circ})} = \cos{(-70^{\circ})}$
Recall further that:
Cosine is an even function so $\cos{(-\theta)} = \cos{\theta}$.
Thus,
$\cos{(-70^{\circ})} = \cos{70^{\circ}}$
And so
$\dfrac{\sin{70^{\circ}}}{\cos{(-430^{\circ})}} = \dfrac{\sin{70^{\circ}}}{\cos{70^{\circ}}}$
Since $\tan{\theta} =\dfrac{\sin{\theta}}{\cos{\theta}}$, then
$\dfrac{\sin{70^{\circ}}}{\cos{70^{\circ}}} = \tan{70^{\circ}}$
Recall that $\tan{(-\theta)}= - \tan{\theta}$.
Hence,
$\tan{(-70^{\circ})} = - \tan{70^{\circ}}$
and
$\dfrac{\sin{70^{\circ}}}{\cos{70^{\circ}}} + \tan{(-70^{\circ})} = \tan{70^{\circ}} - \tan{70^{\circ}} = \boxed{0}$
