Let $x$ be a real number such that $\sec x - \tan x = 2.$  Find $\sec x + \tan x.$
I notice that this problem involves secant and tangent, which are trigonometric functions that are related to cosine and sine. One identity that comes to mind is $\sec^2 x - \tan^2 x = 1,$ which is equivalent to $(\sec x - \tan x)(\sec x + \tan x) = 1.$ This looks promising, because it has both $\sec x - \tan x$ and $\sec x + \tan x$ in it. If I can use this identity to solve for $\sec x + \tan x,$ then I would be done. To do that, I need to plug in the given value of $\sec x - \tan x,$ which is 2. So I get $(2)(\sec x + \tan x) = 1,$ and dividing both sides by 2, I get $\sec x + \tan x = \frac{1}{2}.$ 