Question

How do you evaluate `arctan ( - (sqrt 3))`?

Answer 1

`arctan(-sqrt3)=-pi/3`.

Explanation:

`arctanx=theta, x in RR iff tantheta=x, theta in (-pi/2,pi/2)`

We have, `tan(-pi/3)=-tan(pi/3)=-sqrt3, where, -pi/3 in (-pi/2,pi/2)`

`:. arctan(-sqrt3)=-pi/3`.