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):}#
