Answer
$\cos \left( {{\tan }^{-1}}x \right)=\frac{1}{\sqrt{1+{{x}^{2}}}}$.
Work Step by Step
Let $\left( {{\tan }^{-1}}x \right)=y$.
Thus
$\left( \tan y \right)=x$
For $0\le x\le 1$, $0\le y\le \frac{\pi }{4}$.
Here, $\left( \tan y \right)=x$ is the ratio of the perpendicular line to the base.
Therefore, the perpendicular is $x$ and the base is $1$.
Let the third side of the triangle that is hypotenuse be $a$.
Now, by using the Pythagorean theorem, we get;
$a=\sqrt{{{x}^{2}}+1}$
Since $\left( {{\tan }^{-1}}x \right)=y$, therefore $\cos \left( {{\tan }^{-1}}x \right)$ becomes $\cos y$.
Now we will take cos as the ratio of the base to the hypotenuse.
Therefore
$\cos y=\frac{1}{\sqrt{1+{{x}^{2}}}}$
Which is $\cos \left( {{\tan }^{-1}}x \right)=\frac{1}{\sqrt{1+{{x}^{2}}}}$.
