How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of 
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find 
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies 
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?
 A: Consider the following figure:

Triangles $\triangle ABC$ and $\triangle ACD$ are right triangles, and $AB\cong BD$, so $\triangle ABD$ is isosceles with 
$$\angle BAD=\angle BDA={1\over2}(\pi-\angle ABD)={1\over2}\left(\pi-\left({\pi\over2}-\angle BAC \right)\right)={\pi\over4}+{1\over2}\angle BAC$$
where all angles are taken in a positive sense.  We also have
$$\angle CAD=\angle BAD-\angle BAC={\pi\over4}-{1\over2}\angle BAC$$
Finally, taking care with minus signs, we have
$$\angle CAD=-\tan^{-1}(x-\sqrt{1+x^2})\quad\text{and}\quad\angle BAC=\tan^{-1}x$$
which gives the desired identity
$$\tan^{-1}(x-\sqrt{1+x^2})={1\over2}\tan^{-1}x-{\pi\over4}$$
Remark (added later):  The picture and proof here assume $x\ge0$.  To cover the case $x\lt0$ requires either its own picture or an appeal to some kind of analytic continuity.
A: By virtue of solving for $x$, the original equation $$y = \tan^{-1} (x - \sqrt{1+x^2})$$ implies $$x = \frac{1}{2}(\tan y - \cot y) = \frac{\sin^2 y - \cos^2 y}{2 \sin y \cos y} = -\cot 2y,$$ hence $$y = \frac{1}{2}\cot^{-1} (-x) = -\frac{1}{2} \cot^{-1} x.$$  Now recalling that $$\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2},$$ we readily obtain $$y = \frac{1}{2} \tan^{-1} x - \frac{\pi}{4}.$$
A: Let me rewrite the proof a bit differently.
We are required to prove (RTP) that $$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\frac{\pi}{2}$$
Proof:
Plan: We will use the formula F first, followed by the identity I.
Let us call the following identity I:
$\tan^{-1}(x) + \tan^{-1} (1/x) =$
\begin{cases}
\hfill \pi/2 & \text{if $x\geq 0$}\\
      -\pi/2 & \text{if $x<0$}
\end{cases}
Next, call the following formula F:
$2 \tan^{-1} (x) =$
\begin{cases}
\hfill \tan^{-1} \frac{2x}{1-x^2} & \text{if $|x|\neq 1$}\\ 
       \frac{\pi}{2} & \text{if $x = 1$}\\
       -\frac{\pi}{2} & \text{if $x = -1$}
\end{cases}
Also note that $x - \sqrt{1+x^2} < 0$, 
therefore $|x - \sqrt{1+x^2}| > 1$ is equivalent to  $x - \sqrt{1+x^2} < -1 \implies x < 0$
Done with the ground work!
If $y < 0$, 
LHS of RTP =
$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$
[we used formula F (first case) first, followed by identity I (second case)]
If $y > 0$, 
LHS of RTP =
$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$
[we used formula F (first case) first, followed by identity I (first case)]
The case $y = 0$ is obvious!
A: First let's figure out what $C$ is. Letting $x=0$, we see that $\arctan(-1) = C$ (i.e. $C = -\frac{\pi}{4}$ or $C = \frac{3\pi}{4}$). Since inverse trigonometric functions are hard to work with, let's take $\tan$ of both sides. Doing so we get
$$\tan\left(\arctan(x-\sqrt{1+x^2})\right) = \tan\left(\frac{1}{2}\arctan(x)+C\right).$$
The left hand side is nothing more than $x-\sqrt{1+x^2}$. As for the right hand side, we can employ the tangent sum formula:
$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}.$$
Thus
$$\tan\left(\frac{1}{2}\arctan(x)+C\right) = \frac{\tan\left(\frac{1}{2}\arctan(x)\right) + \tan(\arctan(-1))}{1-\tan\left(\frac{1}{2}\arctan(x)\right)\tan(\arctan(-1))}.$$
We have that $\tan\left(\frac{1}{2}\arctan(x)\right) = \frac{\sqrt{1+x^2}-1}{x}$ which can be verified by half angle formulas. If you put these pieces together and do some algebra, you will have your answer.
A: Let $x=\tan\theta$.  Then what I like to think of as the  “cartographer's” tangent half-angle formula says
$$
\tan\theta \pm \sec\theta = \tan\left(\frac\pi4 \pm \frac \theta 2\right).
$$
If $\sec\theta>0$ then $\sec\theta=\sqrt{1+x^2}$.
A: First, you have $\arctan (\tan y)=y$ for $y\in]-\pi/2, \pi/2[$.
And for all $y\neq \pi/4+k\pi/2$,
$$\tan (2y)=\frac{2\tan y}{1-\tan^2 y}$$
Thus, for all $z\in]-1,1[$, $y=\arctan z$ is in $]-\pi/4,\pi/4[$, and you can apply the above:
$$\tan(2\arctan z)=\frac{2z}{1-z^2}$$
And since $2\arctan z \in ]-\pi/2,\pi/2[$, you can also apply the first identity to get
$$\arctan \tan (2\arctan z)=2\arctan z=\arctan \left(\frac{2z}{1-z^2}\right)$$
Bear in mind that this is proved only for $z\in]-1,1[$, and it actually is wrong outside this interval.
Now, for $x>0$, $x-\sqrt{1+x^2}<0$ since $1+x^2>x^2$ implies $\sqrt{1+x^2}>x$.
But it's also $>-1$, since
$$(1+x)^2=x^2+1+2x>x^2+1$$
And taking square root, $1+x>\sqrt{1+x^2}$
Thus, for $x>0$, $-1<x-\sqrt{1+x^2}<0$.
It's thus a good candidate to apply the formula above with arctangent, which is true for $|z|<1$:
That is, for $x>0$,
$$2\arctan (x-\sqrt{1+x^2})=\arctan \left(\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}\right)$$
And
$$\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}=\frac{2(x-\sqrt{1+x^2})}{-2x(x-\sqrt{1+x^2})}=-\frac{1}{x}$$
Thus, still for $x>0$,
$$2\arctan (x-\sqrt{1+x^2})=-\arctan\frac{1}{x}$$
And since $\arctan x+\arctan \frac1x=\pi/2$ for $x>0$, you have
$$2\arctan (x-\sqrt{1+x^2})=\arctan (x) -\pi/2$$

Now, we have to prove the same for $x<0$.
Then, $x-\sqrt{1+x^2}<-1$, so we can't apply our formula with arctangent.
However, its inverse is in $]-1,0[$, and for $x<0$,
$$\arctan (x-\sqrt{1+x^2})+\arctan \left(\frac{1}{x-\sqrt{1+x^2}}\right)=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})-\arctan (x+\sqrt{1+x^2})=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})=\arctan (x+\sqrt{1+x^2})-\pi/2$$
Then, still for $x<0$, $x+\sqrt{1+x^2} \in ]0,1[$ and
$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{1-(x+\sqrt{1+x^2})^2}$$
$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{-2x(x+\sqrt{1+x^2})}=-\arctan \frac1x$$
And $\arctan x+\arctan \frac1x=-\pi/2$ (remember that $x<0$), hence
$$2\arctan (x+\sqrt{1+x^2})=\arctan (x)+\pi/2$$
And finally
$$2\arctan (x-\sqrt{1+x^2})=2\arctan (x+\sqrt{1+x^2})-\pi=\arctan (x)-\pi/2$$
This formula is now proved for $x>0$ and for $x<0$. It's also true for $x=0$, since then it amounts to $2\arctan (-1)=-\pi/2$.
