#x = sec theta + tan theta#
#x = { 1 + sin theta}/ cos theta#
All the answers are of the form # {x^2 \pm 1}/{kx}# so let's square # x#:
# x ^2 = {1 + 2 sin theta + sin ^2 theta}/{cos ^2 theta}#
# x ^2 = {1 + 2 sin theta + sin ^2 theta}/{1 - sin ^2 theta}#
Let #s = sin theta#
#x^2 - x^2 s^2 = 1 + 2s + s^2#
#(1 + x^2)s^2 + 2s + (1-x^2) = 0#
That factors!
# (s + 1) ((1+ x^2 )s + (1- x^2 )) = 0#
# s= -1 or s = {1-x^2}/{1+x^2} #
#sin theta=-1# means #theta=-90^circ# so the cosine is zero and #sec theta + tan theta# is undefined. So we can ignore that and conclude
#sin theta = {1-x^2}/{1+x^2} #
That's a right triangle whose remaining side is
# \sqrt{ (1+x^2)^2 - (1-x^2)^2 } = sqrt{2(2x^2)} = |2x|#
So
#tan theta = pm {1-x^2}/{2x} #
We could worry about the absolute value, but let's just call this choice #D.#