Question

Evaluate the following:

`cot^-1(cot  (19pi)/6)`

Answer 1

We know that

cot^-1 (cot θ) = θ,   (0, π)

We have

`cot^-1(cot  (19pi)/6)=cot^-1[cot(pi+pi/6)]`

`=cot^-1(cot  pi/6)`

`=pi/6`