TITLE: Inverse of $\tan^{2}\theta$?
QUESTION [1 upvotes]: I re-arranged: $$3\tan^{2} x -1=0$$ to get $\tan^{2}\theta = \frac{1}3$. I noticed the inverse of the $cos, sin$ and $tan$ functions are written as $\cos^{-1}\theta,  \sin^{-1}\theta$ and $\tan^{-1}\theta$ respectively, does this mean the inverse of $\tan^{2}\theta$ would equal $\tan^{(2-1=1)}\theta = \tan\theta$ ? Also is it referred to Arc-$function$ or the inverse of the function, I've heard they're two different things but the distinction is ambiguous to me.

REPLY [1 votes]: REMEMBER: $tan^2\;x$ is a simplification of $(tan(x))^2$.
It's easier than it seems, root both sides so $tan(x) = \frac{\pm 1}{\sqrt3}$
Now inverse tan $\frac{1}{\sqrt{3}} $ ...  $tan^{-1}(\frac{1}{\sqrt{3}})$
and you get: $\theta = 30$ this is the principal value (closest to the origin); you can find the limitless other solutions by $\pm 180$
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