Given the identity sec^2(theta) = tan^2(theta)+1 we can take the square root of both sides of the equation to find sec(theta) in terms of tangent. But we must account for the sign of the function.
Noting that the secant function has a period of 2pi we have
sec(theta) > 0 for theta in (-pi/2 + 2pik, pi/2+2pik), k in ZZ
sec(theta) < 0 for theta in (pi/2 + 2pik, (3pi)/2 + 2pik), k in ZZ
(If we look from the standpoint of the unit circle, then the above means that the secant function is positive in quadrants I and IV and negative in quadrants II and III)
So, after taking the square root, we have
sec(theta) = {(sqrt(tan^2(theta)+1)" "-pi/2+2pik < theta < pi/2 + 2pik), (-sqrt(tan^2(theta)+1)" "pi/2+2pik < theta < (3pi)/2 + 2pik):}
Thus
sec(theta)/7 = {(sqrt(tan^2(theta)+1)/7" "-pi/2+2pik < theta < pi/2 + 2pik), (-sqrt(tan^2(theta)+1)/7" "pi/2+2pik < theta < (3pi)/2 + 2pik):}
