Question

How do you evaluate `cot^-1(1/sqrt3)` without a calculator?

Answer 1

`"arc"cot(1/sqrt3)=1/3pi`

Explanation:

Let `theta="arc"cot(1/sqrt3)`

`cottheta=1/sqrt3`

`tantheta=sqrt3`

This is one of the angles you should know and works out to be `1/3pi`