Question

How do you find the value for `tan^-1[tan(5pi/7)]`?

Answer 1

The formula
`tan^(-1)[tan(a)]=a`
works for `a in (-pi/2;pi/2)`

In our case `(5pi)/7>pi/2` but we can use the periodicity of `tan`:
`tan((5pi)/7)=tan((5pi)/7-pi)=tan(-(2pi)/7)`
`tan^(-1)[tan((5pi)/7)]=tan^(-1)[tan(-(2pi)/7)]=-(2pi)/7`

Btw, a similar formula
`tan[tan^(-1)(a)]=a`
works for all `a in RR`.