Let $x$ be a real number such that $\sec x - \tan x = 2.$  Find $\sec x + \tan x.$
We have that
\[(\sec x - \tan x)(\sec x + \tan x) = \sec^2 x - \tan^2 x = \frac{1}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} = \frac{1 - \sin^2 x}{\cos^2 x} = 1,\]so $\sec x + \tan x = \frac{1}{\sec x - \tan x} = \boxed{\frac{1}{2}}.$