Question

prove that cos^2x + sec^2x can never be less than 2 for all values of x

Answer 1

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.)