Answer
$\dfrac{2}{\sqrt {4+x^2}}$
Work Step by Step
The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$
Suppose $ \theta =\tan^{-1} (\dfrac{x}{2})$
$\implies \tan \theta=\dfrac{x}{2}$
Since, $\tan \theta= \dfrac{Opposite}{Adjacent}$
Now, $\cos [\tan^{-1} (\dfrac{x}{2})] =\cos \theta= \dfrac{Adjacent}{hypotenuse}$
Using the Pythagorean Theorem, we get:
$=\dfrac{2}{\sqrt {(2)^2+(x)^2}}=\dfrac{2}{\sqrt {4+x^2}}$
