How do you evaluate $arctan(-1/sqrt3)$ ?

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Answer 1 · 2016-07-11T11:14:23

$arctan(-1/sqrt3)=-pi/6.$

Let $arctan(-1/sqrt3)=theta, theta in (-pi/2,pi/2)$

Then, by defn. of $arctan$ fun, $tantheta=-1/sqrt3 <0,$ so that $theta !in (0,pi/2)$

Now, $tan(-pi/6)=-tan(pi/6)=-1/sqrt3,$ where, $(-pi/6) in (-pi/2,0)$

Thus, $tan(-pi/6)=-1/sqrt3=tantheta,$ and, $tan$ fun. is injective i.e., $1-1$ in $(-pi/2,0)$ , we conclude that $theta=arctan(-1/sqrt3)=-pi/6.$

How do you evaluate arctan(-1/sqrt3)arctan(-1/sqrt3)?