Does $\operatorname{arcsec}(x) = 1 /\arccos(x)$? 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.
 A: 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
$$y=\cos^{-1}\left(\frac{1}{x}\right)$$
A: No. If you graph $\sec^{-1}(x) \cdot \cos^{-1}(x)$, you get:


You can clearly see that it isn't $1$.
A: 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 $$\bbox[4pt, border:1px solid blue;]{\color{red}{\sec^{-1}(x)=\cos^{-1}\left(\frac{1}{x}\right)}}$$
A: 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 $$\boxed{\sec^{-1} x = \cos^{-1}\tfrac1x}$$
A: Isn't. Draw
$$\text{arcsec} x\arccos x$$
A: No, it is false.
Probably you meant $\operatorname{arcsec}(x)=\arccos(1/x)$, which is true.
A: Actually it's: $$\operatorname{arcsec}(x)=\arccos(1/x).$$
A: No, they are not equal. Here is why.
We can look at an algebraic approach to arcsec and arccos first:
Arccos(x) = y
Cos(y) = x
1/sec(y) = x
Sec(y) = 1/x
y = arcsec(1/x)
Now, we have established that
Arccos(x) = y = arcsec(1/x),
Thus, arccos(x) = arcsec(1/x)
And we have got a neat algebraic proof.
Conversely, we can do the reverse by substituting cos for sec in the above equations to get arcsec(x) = arccos(1/x)
Take a graphing calculator, you can substitute the values, and this is the image of the graph
Lastly, the longest side of a triangle is 1/cos(x) = y, and since y is not equal to sec(y), the above equation in the question is wrong ( 1/cos(x) is not sec(y), and 1/arccos(x) is not arcsec(x) ).
