The max value of cos(x)=1 and the min value is -1. Therefore cos2(x)≤1. Let cos2(x)=1-p2 where -1<p<1, for all other values of x where cos2(x)≠1, then sec2(x)=1/(1-p2), so sec2(x)>1.
cos2(x)+sec2(x)=1-p2+1/(1-p2)=(2-2p2+p4)/(1-p2)=2+p4/(1-p2) which is always bigger than 2, because 1-p2>0 since p2<1, and p4 is always positive. (If p=0 cos2(x)+sec2(x)=2 the minimum value.)