As we have seen in another post, identities involving inverse tangent look much less mysterious if analyzed from a purely geometrical perspective. Here you have the chance to further practice on the subject and to demonstrate a more general formula. Let’s start with the relationship

$$\arctan x + \arctan\left(\frac{x+1}{x-1}\right) =\frac{3\pi}{4}, \ \ \mbox{for}\ \ x>1.\tag{1}\label{eq630:1}$$

We will start again with the now ubiquitous right-angled triangle $$\triangle ABC$$ with sides $$\overline{AB} = 1$$, $$\overline{BC} = x$$, and hypothenuse $$\overline{AC} = \sqrt{1+x^2}$$.

1. Extend $$AB$$ with a segment $$\overline{BD} = x$$.
2. Triangle $$\triangle BCD$$ is isosceles. What can you say, then, about the angle $$\gamma = \angle ADC$$?
3. Use what you observed, and the fact that the sum of the interior angles of $$\triangle ACD$$ is $$\pi$$, to write a relationship connecting $$\alpha$$ and $$\beta$$. You should get $\beta = \frac{3\pi}{4}-\alpha.$
4. Draw from $$D$$ the line perpendicular to $$AC$$, that intersects $$AC$$ in $$H$$.
5. $$\triangle ABC$$ and $$\triangle ADH$$ are similar. Why?
6. From the scale factor between these triangles and the fact that $$\overline{AD} = 1+x$$, determine $$\overline{DH}$$, $$\overline{AH}$$, and finally $$\overline{CH} = \overline{AC}-\overline{DH}$$.
7. Write down $$\beta$$ as $$\arctan\left(\frac{\overline{DH}}{\overline{CH}}\right)$$.
8. Use the relationship found in 3. to get your final result.

If you followed without difficulties the previous demonstration, you should now be able to prove also the following identity.

$$\arctan x + \arctan\left(\frac{\sqrt 3 x+1}{x-\sqrt 3}\right)=\frac{5\pi}{6}, \ \ \mbox{for} \ \ x >\sqrt 3.\tag{2}\label{eq630:2}$$

It is just a matter of modifying the measure of the angle $$\gamma$$, and make it equal to $$\frac{\pi}{6}$$. What is then the measure of $$\overline{BD}$$? Recall that $$\frac{\pi}{3}$$ is the measure of the interior angles of an equilateral triangle. Procede from here as you have done before, in order to prove \eqref{eq630:2}.

You might be wondering now whether the expressions \eqref{eq630:1} and \eqref{eq630:2} can be further generalized. After all $$\gamma$$ can be arbitrarily chosen, and in general we would have $$\overline{BD} = \frac{x}{\tan\gamma}$$. Viceversa, if we take $$\overline{BD} = \frac{x}{y}$$, we will get $$\gamma = \arctan y$$.

Use the latter notation and generalize the procedure that you adopted so far. Recalling that
$\beta = \pi – \arctan x – \arctan y,$

obtain the following identity

$\arctan x + \arctan y = \pi – \arctan\left(\frac{x+y}{xy-1}\right),$

which is valid for $$x>\frac{1}{y}>0$$.