I would do the following:
sec(x) + tan(x) = 1
sec(x) = 1 - tan(x)
[sec(x)]^2 = [1 - tan(x)]^2
[sec(x)]^2 = 1 - 2tan(x) + [tan(x)]^2
[sec(x)]^2 - [tan(x)]^2 = 1 - 2tan(x)
But from the identity tan<sup>2</sup>x + 1 = sex<sup>2</sup>x, the left side reduces to 1, thus
1 = 1 - 2tan(x)
0 = -2tan(x)
tan(x) = 0
x = tan<sup>-1</sup>0
Tangent is 0 whenever sine is 0, which is at zero and pi.