<s>[INST] $\arctan(\tan(-\frac{3\cdot \pi}{5}))+\operatorname{arccot}(\cot(-\frac{3\cdot \pi}{5}))=?$ We know that $$\DeclareMathOperator{\arccot}{arccot}\arctan(-x)=-\arctan(x)$$ $$\arccot(-x)=\pi-\arccot(x)$$ $$\cot(-x)=-\cot(x)$$ $$\tan(-x)=-\tan(x)$$So I have figured out that $$\arctan(\tan(-\frac{3\cdot \pi}{5}))=\arctan(-\tan(\frac{3\cdot \pi}{5}))=-\arctan(\tan(\frac{3\cdot \pi}{5}))=-\frac{3\cdot \pi}{5},$$ $$\arccot(\cot(-\frac{3\cdot \pi}{5}))=\arccot(-\cot(-\frac{3\cdot \pi}{5}))=\pi-\arccot(\cot(\frac{3\cdot \pi}{5}))=\pi-\frac{3\cdot \pi}{5}.$$Coming from these, I have solved the problem as follows:$$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\arccot(\cot(-\frac{3\cdot \pi}{5}))=-\frac{3\cdot \pi}{5}+\pi-\frac{3\cdot \pi}{5}=-\frac{6\cdot \pi}{5}
+\pi=-\frac{\pi}{5}.$$But the book says it is incorrect. Could somebody point out where I have made a mistake?
 [/INST] The function $\arctan(\tan x)$ returns a “normalized” $x$, that is the unique angle $x'\in(-\pi/2,\pi/2)$ so that $\tan x=\tan x'$.
Since $-3\pi/5<-\pi/2$, but $2\pi/5=-3\pi/5+\pi\in(-\pi/2,\pi/2)$, you have
$$
\arctan\Bigl(\tan\Bigl(-\frac{3\pi}{5}\Bigr)\Bigr)=\frac{2\pi}{5}
$$
Do similarly for the arccotangent, but with the “normalization” in $(0,\pi)$.
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