Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$ 
Prove that for all $x\in \mathbb R$,  $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$

From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-x-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}=-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}$$I know that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) \rightarrow 0$$But I don't know how to show that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) = 0$$ Can you help me?

Hint:
Let $\arctan x=y,-\pi/2<y<\pi/2,\cos y>0$
$\arccos\dfrac{\tan y}{\sec y}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)=?$
