# How do you show cos (arctan x) = [ 1 / ( sqrt(1 + x^2))]?

Jul 3, 2016

See the Proof given in the following Explanation.

#### Explanation:

Let $\arctan x = \theta \Rightarrow \tan \theta = x , x \in \mathbb{R} , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

Now, ${\sec}^{2} \theta = 1 + {\tan}^{2} \theta = 1 + {x}^{2} \Rightarrow \sec \theta = \pm \sqrt{1 + {x}^{2}}$

$\therefore \cos \theta = \pm \frac{1}{\sqrt{1 + {x}^{2}}}$

But, $\theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) = {Q}_{I V} \cup {Q}_{I} , w h e r e , \cos \theta$ is $+ v e$.

Thus, $\cos \theta = + \frac{1}{\sqrt{1 + {x}^{2}}}$ and, replacing $\theta$ by $\arctan x$, we have proved that,

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