Question

How do you identify the period and asympotes for `y=tantheta`?

Answer 1

`y=tantheta` has a period of `pii` and vertical asymptotes are at `theta=((2n+1)pi)/2`

Explanation:

As `tan(pi+theta)=tantheta`, `y=tantheta` has a period of `pi`.

Further as `theta->((2n+1)pi)/2`, from left or right i.e.

`lim_(theta->(((2n+1)pi)/2)^+)tantheta=oo` and

`lim_(theta->(((2n+1)pi)/2)^-)tantheta=-oo`

Hence vertical asymptotes are at `theta=((2n+1)pi)/2`