Answer
$0$
Work Step by Step
Recall the co-function identity:
$ \sin (\theta)=\cos(90^{\circ}-\theta)$
We know that
$\sin^2 \theta+\cos^2 \theta=1$
Therefore, $1-\cos^2 \ (20^{\circ})-\cos^2 (70^{\circ}) =1-\cos^2 \ (20^{\circ})-\sin^2 (70^{\circ}) \\=1-[\cos^2 \ (20^{\circ})+\sin^2 (20^{\circ})] \\=1-1 \\=0$
