Answer
$\sqrt{\dfrac{x^2}{4}-x+2}$
Work Step by Step
$\theta = tan^{-1} {\dfrac{x-2}{2}}$
$\tan{\theta} = \dfrac{x-2}{2}$
$\sec^2{\theta} = 1+\tan^2{\theta} = 1+(\dfrac{x}{2}-1)^2 =1+\dfrac{x^2}{4}-x+1$
$\sec^2{\theta} = \dfrac{x^2}{4}-x+2$
$\sec{\theta} = \sqrt{\dfrac{x^2}{4}-x+2}$
