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How this equality is verified.

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

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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 $\in\left(0,3\cdot\dfrac\pi2\right)$

See also: Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?

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Use Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

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

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

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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

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

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