$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\operatorname{arccot}(\cot(-\frac{3\cdot \pi}{5}))=?$

Question

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?

Answer 1

For all real $x$ we have $-\frac{\pi}{2}<\arctan{x}<\frac{\pi}{2}$ and $0<\operatorname{arccot}{x}<\pi$.$\DeclareMathOperator{\arccot}{arccot}$

$$\arctan\tan\left(-\frac{3 \pi}{5}\right)+\arccot\cot\left(-\frac{3\pi}{5}\right)=$$ $$=\arctan\tan\frac{2 \pi}{5}+\arccot\cot\frac{2\pi}{5}=\frac{4\pi}{5}$$

Answer 2

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 $(0,\pi)$.