Question

what is the value of y

Answer 1

y=cos^2(x) +sec^2(x)

cos^2(x) + sec^2(x) = cos^2(x) - 2 cos(x) sec(x) + sec^2(x) +2

                                 = [{sec(x) - cos(x)}^2 + 2]......................(equation 1)

since 2 cos(x) sec(x) =2

we know that

[sec(x)-cos(x)]^2 >= 0

now from equation 1

{sec(x) - cos(x)]^2 +2 >=2

hence

y>=2