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Does $\operatorname{arcsec}(x) = 1 /\arccos(x)$? I have looked in a few books and Google'd it but I am not finding my answer.

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No. If you graph $\sec^{-1}(x) \cdot \cos^{-1}(x)$, you get:

-2 to 2 -15 to 15

You can clearly see that it isn't $1$.

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If $\sec^{-1} x = \theta$, then $x = \sec\theta$. This means $\frac1x = \cos\theta$, so $\cos^{-1}\frac1x = \theta$. So your equation is wrong; the correct statement is $$\boxed{\sec^{-1} x = \cos^{-1}\tfrac1x}$$

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  • $\begingroup$ $$(\sec x)^{-1} = (\cos x)^1 \\ \sec^{-1} x = \cos^{-1} x^{-1}$$ it’s kinda satisfying . . . feels like the negative ones cancel each other out $\endgroup$ – gen-z ready to perish Apr 26 '19 at 21:18
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Actually it's: $$\operatorname{arcsec}(x)=\arccos(1/x).$$

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No. It is not. If you look at the definitions

$$y=\frac{1}{\cos x}$$

and then we solve for the x

$$\frac{1}{y}=\cos x$$

$$\cos^{-1}\left(\frac{1}{y}\right) = x$$

and replace $x$ and $y$ to find the inverse

$$y=\cos^{-1}\left(\frac{1}{x}\right)$$

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  • $\begingroup$ Careful, $\cos^{-1}(\cos(x)) \neq x$ in some cases. $\endgroup$ – recursive recursion May 17 '14 at 18:49
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Isn't. Draw $$\text{arcsec} x\arccos x$$

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  • 2
    $\begingroup$ Please avoid link only answers and add a plot if you are telling the OP to test his theory by graphing. $\endgroup$ – Cole Johnson May 16 '14 at 17:03
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No, it is false. Probably you meant $\operatorname{arcsec}(x)=\arccos(1/x)$, which is true.

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It is straight forward, let $$\sec^{-1}(x)=\alpha$$ $$\implies \sec\alpha=x$$ $$\implies \frac{1}{\cos \alpha}=x$$$$\implies \cos \alpha=\frac{1}{x}$$ $$\implies \alpha=\cos^{-1}\left(\frac{1}{x}\right)$$ Substituting the value $\alpha=sec^{-1}(x)$, we get $$\bbox[4pt, border:1px solid blue;]{\color{red}{\sec^{-1}(x)=\cos^{-1}\left(\frac{1}{x}\right)}}$$

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