Is this a correct way to prove that $$ \arctan(\mathrm{e}^{-v}) = \text{arccot}(\mathrm{e}^v) $$ $$\tan x = \frac{1}{\mathrm{e}^v} $$ Turn $\tan x$ into $\sin x$ and $\cos x$ $$\frac{\sin x}{\cos x} = \frac{1}{\mathrm{e}^v}$$ Multiply $\cos x$ $$ \sin x = \frac{\cos x}{\mathrm{e}^v} $$ $$ \mathrm{e}^v \sin x = \cos x$$ $$ \mathrm{e}^v = \frac{\cos x}{\sin x}$$ Multiply by $\mathrm{e}^v$ and divide by $\sin x$ $$ \mathrm{e}^v = \cot x$$ $$\text{arccot}(\mathrm{e}^v) = x$$
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Taking the e-power as it is (replace it by $x$ if you wish), your derivation looks good – imranfat Dec 15 '17 at 03:31
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Looks fine to me :) – Robert Howard Dec 15 '17 at 03:34
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I think it's correct.
Just $$\arctan{x}=\text{arccot}\frac{1}{x}$$ for all $x>0$.
Michael Rozenberg
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Yes but a simpler way to do it is to set everything equal to one another $$\arctan(x)=\frac{1}{e^v} \iff e^v=\mathrm{arccot}(x)$$ $$\frac{\arcsin(x)}{\arccos(x)}=\frac{1}{e^v}\iff e^v=\frac{\arccos(x)}{\arcsin(x)}$$ everything cancels out resulting in that they do equal each other in the set $\{v : v \in \mathbb R\}$ because $\frac{1}{e^v}$ or $e^v$ does not equal zero ever.
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You're abusing the equals sign in the middle of those two lines. It's not true that $\frac{1}{e^v} = e^{v}$. What you mean is $\iff$. And the last line isn't needed. – Matthew Leingang Dec 15 '17 at 03:42
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@Gil, $\arctan x=y\implies x=\tan y$ but not conversely always – lab bhattacharjee Dec 15 '17 at 03:53
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Hint: You may also use the mean value theorem. Let $f(x)=\arctan (e^x)-\arctan (e^{-x})$, so $f'(x)=\displaystyle \frac{e^x}{1+e^{2x}}-\frac{e^{-x}}{1+e^{-2x}}\Longrightarrow f'(x)=\displaystyle \frac{e^x}{1+e^{2x}}-\frac{e^{x}}{1+e^{2x}}=0\Longrightarrow f(x)=$constant by the mean value theorem. If you put $x=0$ you get constant$=0$.
daulomb
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