# $\arctan(2\sqrt2)+2\arctan(\sqrt2)=\pi$ [closed]

How this equality is verified.

$$\arctan(2\sqrt2)+2\arctan(\sqrt2)=\pi$$

Using $$\tan(A+B+C)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{\cdots}$$

$$\tan(\arctan(2\sqrt2)+2\arctan(\sqrt2))=\dfrac{2\sqrt2+\sqrt2+\sqrt2-2\sqrt2(\sqrt2)^2}{\cdots}=0$$

$$\implies\arctan(2\sqrt2)+2\arctan(\sqrt2)$$ must be multiple of $$\pi$$

Again, as $$2\sqrt2,\sqrt2>0$$

$$\dfrac\pi2>\arctan(2\sqrt2)>\arctan(\sqrt2)>0$$

$$\implies0<\arctan(2\sqrt2)+2\arctan(\sqrt2)<3\cdot\dfrac\pi2$$

Now what is the multiple of $$\pi$$ that lies $$\in\left(0,3\cdot\dfrac\pi2\right)$$

$$2\arctan\sqrt2=\pi+\arctan\dfrac{2\sqrt2}{1-(\sqrt2)^2}=?$$

Now $$\arctan(-x)=-\arctan x$$

We can use the sum identity for the tangent function

$$\tan(x\pm y)=\frac{\tan x\pm \tan y}{1\mp \tan x \tan y}$$

to check the given identity by

$$\arctan(2\sqrt2)+2\arctan(\sqrt2)=\pi$$

$$\iff \arctan(2\sqrt2)+\arctan(\sqrt2)=\pi-\arctan(\sqrt2)$$

that is

$$\frac{2\sqrt 2+\sqrt 2}{1-2\sqrt 2 \cdot \sqrt 2}\stackrel{\text{?}}=\frac{0-\sqrt 2}{1+0}$$