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I was curious if one can bring the negative sign out as we do in case of $\sin^{-1}(x)$, $\tan^{-1}(x)$, & $\csc^{-1}(x)$. Can one do the same for $\cos^{-1}(x)$, $\sec^{-1}(x)$, & $\cot^{-1}(x)$? For example,

$$\cot^{-1}(-\tan(x))$$

$$=-\cot^{-1}(\tan(x))...(i)$$

Can I write (i)? I'm asking this because if $x=45^{\circ}\text{(any acute angle)}$,

$$\cot^{-1}(-\tan(45^{\circ}))$$

$$=-\cot^{-1}(\tan(45^{\circ}))$$

$$=-\cot^{-1}(1)$$

$$=-45^{\circ}$$

Now, $-45^{\circ}$ is outside the restricted range of $\cot^{-1}(x)$:$(0,\pi)$. So, is (i) valid to write?

tryingtobeastoic
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2 Answers2

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The equality $(i)$ is not correct. Since $\operatorname{arccot}$ is the inverse of the restriction of $\cot$ to $(0,\pi)$, it is a function from $\Bbb R$ into $(0,\pi)$ and therefore it cannot possibly be an odd function. In fact, if $x\in\left(-\frac\pi2,\frac\pi2\right)$, then $\operatorname{arccot}\bigl(\tan(x)\bigr)=x+\frac\pi2$.

José Carlos Santos
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It depends: step (i) is valid if and only if you use the first instead of the second definition of $\textrm{arccot}$ here. This is because $\textrm{arccot}$ is an odd function only in the first definition.

  1. Wolframalpha uses the first definition, so it does give the answer that you suggested, $-45^\circ.$

  2. Desmos uses the second definition (which is my preference), so it instead gives $135^\circ.$ The link shared by Henry in the above comments, as well as José's answer, corresponds to this.

P.S. On the other hand, it is always valid to write $$\arctan(-x)=-\arctan (x).$$

ryang
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