Can $\sec x $ be not only $1/\cos x $ but also $\cos(1/x)$?

Question

The question was finding the trig value of $\sec^{-1}(-2/\sqrt{3})$. And, my teacher solved it making it to $\cos^{-1}(-\sqrt{3}/2)$. Is that possible? Can $\sec^{-1} x$ be $\cos^{-1}(1/x)$? ( I'm sorry I'm not handy with the signs. The $-1$ means the inverse function and the things next to it are the angles.)

Answer 1

Let $\sec^{-1}(x)=y$

Then $\sec(y)=x$

Which means $\cos(y)=\frac{1}{x}$

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

Therefore $\sec^{-1}(x)=\cos^{-1}(\frac{1}{x})$

Answer 2

It is true that the $\sec^{-1}x = \cos^{-1}\left(1/x\right)$. Let $\theta$ be the $\sec^{-1}x$. That means $\sec \theta = x$. What then is the cosine? It's $1/x$