Can we always write tan^-1 (x) as cot^-1(1/x),sin^-1(x) as cosec^-1(x) and sec^-1(x) as cos^-1 (x) or are there any restrictions on this fact because of domain ?
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1To be clear... when you use $~^{-1}$ are you talking about functional inverse or multiplicative inverse? E.g. when you say $\sin^{-1}(x)$ do you mean $\arcsin(x)$ or do you mean $\frac{1}{\sin(x)}$? – JMoravitz Aug 21 '19 at 12:55
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Is the logic here $$ \tan{y} = x \qquad \Leftrightarrow \qquad \frac{1}{\tan{x}} =\text{cot}{x} = \frac{1}{y} $$ ? – Matti P. Aug 21 '19 at 12:56
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I mean Arc sin x – Schwarz Kugelblitz Aug 21 '19 at 12:56
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@MattiP. you seem to have mixed up your y's and x's. – JMoravitz Aug 21 '19 at 12:57
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How have I mixed up my y's and X's?...I can't read what you have written mathematically.. – Schwarz Kugelblitz Aug 21 '19 at 12:59
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I made that message to MattiP, not to you Schwarz. Matti had written $\tan (\color{red}{y})=x \iff \frac{1}{\tan(\color{red}{x})}=\frac{1}{y}$. The variables in red should have matched – JMoravitz Aug 21 '19 at 13:00
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In any case., yes, $\arctan(x)$ is the same as $\text{arccot}(\frac{1}{x})$ almost everywhere, and similarly for the others. The places where it doesn't match would be due to division by zero errors. – JMoravitz Aug 21 '19 at 13:03
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Yes, I made a typo there. Oopsie. – Matti P. Aug 21 '19 at 13:05
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The principle values for the inverse functions were chosen exactly with the intent that $$\operatorname{Arcsin}\left(\frac 1x\right) = \operatorname{Arccsc}(x),\quad \operatorname{Arccsc}\left(\frac 1x\right) = \operatorname{Arcsin}(x)$$ $$\operatorname{Arccos}\left(\frac 1x\right) = \operatorname{Arcsec}(x),\quad \operatorname{Arcsec}\left(\frac 1x\right) = \operatorname{Arccos}(x)$$ everywhere in their domains except for $x = 0$.
However, it is common to take $\left(-\frac \pi 2, \frac \pi 2\right)$ as the range of $\operatorname{Arctan}$, but $(0,\pi)$ as the range of $\operatorname{Arccot}$. Therefore
$$\operatorname{Arctan}\left(\frac 1x\right) = \operatorname{Arccot}(x),\quad \operatorname{Arccot}\left(\frac 1x\right) = \operatorname{Arctan}(x)$$ holds for $x > 0$, but $$\operatorname{Arctan}\left(\frac 1x\right) = \operatorname{Arccot}(x) - \pi,\quad \operatorname{Arccot}\left(\frac 1x\right) = \operatorname{Arctan}(x) + \pi$$ holds for $x < 0$.
Paul Sinclair
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