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?
 A: An easy method is to consider the function $f(x)=\arctan{x}+\arccos{\frac{x}{\sqrt{x^2+1}}}$ then show that $f’(x)=0$ and deduce that $f(x)$ is a constant function and therefore $f(x)=f(0)=\pi/2$
A: Draw a right triangle whose tangent is $x$.  Label the opposite side $x$ and the other leg $1$.  Now find the length of the hypotenuse; it's $\sqrt{x^2 + 1}$.  
Then, compute the cosine of the complementary angle.  You are done.
A: Hint: The classic way to solve these type of function is to compute the derivative of $arctan(x)$ and ${\pi\over 2}-arccos({x\over\sqrt{1+x^2}})$ and show that they are equal, now both functions coincide at $0$.
A: Hint:
Let $\arctan x=y,-\pi/2<y<\pi/2,\cos y>0$
$\arccos\dfrac{\tan y}{\sec y}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)=?$
A: By definition, $\arctan x=\theta$ means
$$\tan\theta=x\quad\hbox{and}\quad -\frac\pi2<\theta<\frac\pi2\ .$$
Let $\theta$ be the RHS of your equation.  We have
$$-1<\frac{x}{\sqrt{1+x^2}}<1\quad
  \Rightarrow\quad 0<\arccos\frac{x}{\sqrt{1+x^2}}<\pi\quad
  \Rightarrow\quad -\frac\pi2<\theta<\frac\pi2\ ,$$
so the second part of the condition is true.  The fact that $0<\arccos\frac{x}{\sqrt{1+x^2}}<\pi$ also means that
$$\sin\Bigl(\arccos\frac{x}{\sqrt{1+x^2}}\Bigr)
  =\sqrt{1-\cos^2\Bigl(\arccos\frac{x}{\sqrt{1+x^2}}\Bigr)}$$
and not the negative of this square root, so
$$\eqalign{\tan\theta
  &=\cot\arccos\frac{x}{\sqrt{1+x^2}}\cr
  &=\frac{\cos(\arccos(x/\sqrt{1+x^2}))}{\sin(\arccos(x/\sqrt{1+x^2}))}\cr
  &=\frac{x/\sqrt{1+x^2}}{\sqrt{1-\frac{x^2}{1+x^2}}}\cr
  &=x\ .\cr}$$
