Find `tan^(-1) tan ((2pi)/3)`

Explanation

(i) `tan^(-1) (tan.(2pi)/(3)) != (2pi)/(3), " as " (2pi)/(3)` does not lie between `- (pi)/(2) and (pi)/(2)`
Now, `tan^(-1) (tan.(2pi)/(3)) = tan^(-1) (tan(pi - (pi)/(3)))`
`= tan^(-1) (- tan. (pi)/(3))`
`= tan^(-1) (tan(-(pi)/(3)))`
`= -(pi)/(3)`
(ii) `tan^(-1) (tan(-6)) = tan^(-1) (tan(2 pi -6))`
`= 2pi - 6 " as " (2pi - 6) in (-pi//2, pi//2)`