Answer
$0$
Work Step by Step
We know that $\cos{(\theta)}=\sin{(90^{o}-\theta)}$, and $\tan{(\theta})=\frac{\sin{(\theta)}}{\cos{(\theta)}}$, hence $\tan{(20^{o})}-\frac{\cos{(70^{o})}}{\cos{(20^{o})}}=\tan({20^{o}})-\frac{\sin{(20^{o})}}{\cos{(20^{o})}}=\tan{(20^{o})}-\tan{(20^{o})}=0.$
