How to evaluate $\arctan(1/\tan(-x))$?

Question

How to evaluate the value of $$\arctan\left(\frac{1}{\tan(-x)}\right)$$ by using the expression of $x$?

Answer 1

HINT

Recall that by trigonometric identities

$$\arctan u + \arctan \dfrac{1}{u}= \begin{cases} \dfrac \pi 2, \text{if }u > 0 \\ - \dfrac \pi 2, \text{if }u < 0 \end{cases}$$

therefore

• for $\tan(-x)>0$

$$\arctan\left(\frac{1}{\tan(-x)}\right) = \dfrac \pi 2-\arctan(\tan(-x))$$

• for $\tan(-x)<0$

$$\arctan\left(\frac{1}{\tan(-x)}\right) = -\dfrac \pi 2-\arctan(\tan(-x))$$

Also recall that

$$\arctan (\tan y)=y \iff -\dfrac \pi 2<y<\dfrac \pi 2$$

otherwise we need to add some integer multiple of $\pi$.

Answer 2

$\arctan(\cot(-x))=\arctan\left(\tan\left(\dfrac\pi2+x\right)\right)=n\pi+\dfrac\pi2+x$

where $n$ is an integer such that $$-\dfrac\pi2<n\pi+\dfrac\pi2+x<\dfrac\pi2$$

$\cot(-x)$ can also be written as $-\cot x=\tan\left(x-\dfrac\pi2\right)$

Answer 3

If $y=\arctan\frac1{\tan(-x)}$, then $\tan y=\frac 1{\tan(-x)}=\cot(-x)=\tan(\frac\pi2-(-x))$, which suggests that $y=x+\frac\pi2$ up to a multiple of $\pi$. We conclude that $$\arctan\frac1{\tan(-x)}=x+\frac\pi2+k\pi$$ where $k$ is the unique integer (if it exists) that makes this value $\in(-\frac\pi2,\frac\pi2)$.