How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly?

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.

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 $$y = \frac{1}{2} \tan^{-1} x - \frac{\pi}{4}.$$

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!