Question

How do you evaluate `tan^-1(tan((13pi)/6))`?

Answer 1

`pi/6`.

Explanation:

`tan^-1{tan((13pi)/6)}=tan^-1{tan((12pi+pi)/6)}`

`=tan^-1{tan(2pi+(pi)/6)}`

`=tan^-1{tan((pi)/6)}...[because, tan(2pi+x)=tanx]`

`=pi/6`

In the last step, we have used the following Defn. of `tan^-1` fun.:

`theta=tan^-1x, x in RR iff tantheta=x, theta in (-pi/2,pi/2).`