Geometric justification of the trigonometric identity $\arctan x + \arctan \frac{1-x}{1+x} = \frac \pi 4$

Solution 1:

In answer to the first question, I have been trying to upload a picture but without success, so I'll have to describe it instead.

1. Draw an isosceles right-angled triangle OAC with A on the x axis and C on the y axis, and O as the origin.

2. Choose a point P on the line AC, with Q as the foot of the perpendicular from P onto OA

3. Let B be the point on OA so that Q is the midpoint of BA

4. Let OQ be 1 unit long, and PQ be of length $x$, in which case BQ and AQ also have this length

5. We now have $\tan POQ=x$ and $\tan OCB=\frac{1-x}{1+x}$

6. It simply remains to show that angle BCP = angle POQ, which is easy enough since triangle CPB is right angled and CP has length $\sqrt2$

I hope this is clear enough

Solution 2:

Edit. Here's a simpler trigonograph that's also purer, in the sense that it avoids auxiliary proportions in favor of relating elements in the identity more directly. It therefore requires less (or perhaps no) explanation.

$$45^\circ = \alpha + \beta = \operatorname{atan}\frac{x}{1} + \operatorname{atan}\frac{1-x}{1+x} \qquad (0 \leq x \leq 1)$$

Interestingly, this diagram has the same basic structure as my Angle-Sum and -Difference trigonographs, as well as the one for $p\sin\theta + q \cos\theta$.

My previous answer:

$$\tan \theta = \frac{x}{1} \qquad \tan \phi = \frac{|\overline{XY}|}{|\overline{XZ}|} = \frac{|\overline{XP}|}{|\overline{XQ}|} = \frac{1-x}{1+x}$$

$$\implies\qquad \operatorname{atan} x + \operatorname{atan}\frac{1-x}{1+x} = \theta + \phi = \frac{\pi}{4}$$

Solution 3:

Here's an attempt to address @Michael's desire for a solution that treats the two angles symmetrically.

It's based on this preliminary result:

Lemma. If $Q$ is the orthocenter of acute $\triangle ABC$, then $$\tan \angle ABQ = \frac{|\overline{AQ}|}{|\overline{BC}|}$$

(Proof is left as an exercise to the reader. Hint: Note the congruent angles at $A$ and $C$.)

With that, we can construct the following:

$$\tan \alpha = \frac{x\sqrt{2}}{\sqrt{2}} = x \qquad \tan \beta = \frac{1-x}{1+x} \quad\implies\quad \operatorname{atan}x + \operatorname{atan}\frac{1-x}{1+x} = \alpha+\beta = \frac{\pi}{4}$$