Answer
$\cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1+x^2}}$
Work Step by Step
We want to find $\cos(\tan^{-1}(x))$.
We can construct a right triangle with legs of lengths $1$ and $x$ and a hypotenuse of $\sqrt{1+x^2}$. Since we have a $\tan^{-1}(x)$ term, let the angle between the side length of $1$ and hypotenuse by $\theta$. That means $\tan(\theta) = x$, which gives $\theta = \tan^{-1}(x)$. Looking at our original equation, we can use the substitution $\theta = \tan^{-1}(x)$ and just look to find $\cos(\theta)$. From our right triangle, $\cos(\theta) = \frac{Adjacent Side}{Hypotenuse} = \frac{1}{\sqrt{1+x^2}}$.
