sec(α)² = 1+tan(α)²
sec(α) = √(1+tan(α)²)
You have α = arctan(x/2), so tan(α) = x/2.
sec(α) = √(1+(x/2)²)
sec(arctan(x/2)) = (1/2)√(x²+4)
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Using SOH CAH TOA, you can draw a right triangle with adjacent side 2 and opposite side x. The angle of interest will then have a tangent equal to O/A = x/2. The secant is the reciprocal of the cosine, so is H/A. You can find H using the Pythagorean theorem. It is √(2²+x²), so you have
... secant(arctan(x/2)) = (√(4+x²))/2 . . . . as above