Help to simplify $\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)$ Can someone help me simplify the argument of $\arctan$ in this problem ?
$$\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)$$
 A: It might be helpful to take the inverse of this function by solving for $x$.
We have the following:
$$y=\arctan\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$
Take the $\tan$ of both sides:
$$\tan y=\frac{\sqrt{1+x^2}-1}{x}$$
Muultiply both sides by $x$ and add $1$:
$$x\tan y+1=\sqrt{1+x^2}$$
Square both sides:
$$x^2\tan^2 y+2x\tan y+1=1+x^2$$
Subtract both sides by $1+x^2$:
$$x^2(\tan^2 y-1)+2x\tan y=0$$
Now, clearly, the original function is undefined for $x=0$, so we can conclude that $x \neq 0$. Therefore, divide by $x$:
$$x(\tan^2 y-1)+2\tan y=0$$
Subtract both sides by $2\tan y$ and divide by $\tan^2 y-1$:
$$x=-\frac{2\tan y}{\tan^2 y-1}$$
Now, this looks very similar to the $\tan(2y)$ identity. However, the denominator is flipped and there is a $-$ sign out front. Therefore, distribute the negative across the denominator:
$$x=\frac{2\tan y}{1-\tan^2 y}=\tan 2y$$
Now, we can solve this equation for $y$ to get a much simpler form of our original equation. Take the $\arctan$ of both sides:
$$\arctan x=2y$$
Divide both sides by $2$ and switch the sides of the equation:
$$y=\frac{\arctan x}{2}$$
A: Take $x=\tan \theta$ So it becomes ${\sec\theta -1\over tan \theta}$ multiply up and down by $\cos\theta$. Becomes ${1-cos\theta\over sin \theta}={2\sin^2(\theta/2)\over 2\sin(\theta/2)\cos (\theta/2)}=\tan(\theta/2)$
So the answer is $\theta/2 = (tan^{-1}x)/2$
A: Differentiate with respect to $x$:
$$
\frac{\partial}{\partial x}\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)=\frac{\frac{1}{x^2+\sqrt{x^2+1}+1}}{1+\left(\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)\right)^2}=\frac{1}{2(1+x^2)}
$$
Thus:
$$
\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)=\frac{1}{2}\arctan(x)+C
$$
Since both sides are $0$ at $x=0$, then  $C=0$.
A: There's a geometric reason for why you get $\frac12 \arctan x$. Write your expression as $\arctan\frac{x}{\sqrt{1+x^2}+1}$ and look at the pictures in this answer (about taking square roots of complex numbers) for the number $z=1+ix$, so that $\phi=\arctan x$ and $r = \sqrt{1+x^2}$, and note that $w=(x+\sqrt{1+x^2})+ix$ so that your expression is the argument of $w$.
