## Precalculus (6th Edition) Blitzer

$\cos \left( {{\tan }^{-1}}x \right)=\frac{1}{\sqrt{1+{{x}^{2}}}}$.
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 $\cos \left( {{\tan }^{-1}}x \right)=\frac{1}{\sqrt{1+{{x}^{2}}}}$.