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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$\begingroup$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 $$\boxed{\sec^{-1} x = \cos^{-1}\tfrac1x}$$
$\endgroup$ 1 $\begingroup$Actually it's: $$\operatorname{arcsec}(x)=\arccos(1/x).$$
$\endgroup$ $\begingroup$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
$$y=\cos^{-1}\left(\frac{1}{x}\right)$$
$\endgroup$ 1 $\begingroup$No. If you graph $\sec^{-1}(x) \cdot \cos^{-1}(x)$, you get:
You can clearly see that it isn't $1$.
$\endgroup$ $\begingroup$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 $$\bbox[4pt, border:1px solid blue;]{\color{red}{\sec^{-1}(x)=\cos^{-1}\left(\frac{1}{x}\right)}}$$
$\endgroup$ $\begingroup$Isn't. Draw$$\text{arcsec} x\arccos x$$
$\endgroup$ 1 $\begingroup$No, it is false. Probably you meant $\operatorname{arcsec}(x)=\arccos(1/x)$, which is true.
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