You need to notice that the function does not exist at `x = k*pi ` since `tan(k*pi) = 0` , hence, there exists vertical asymptotes at `x = k*pi, k in Z.`

You need to evaluate the critical points of the function, if any, such that:

`f'(x) = (1'*tan x - 1*(tan x)')/(tan^2 x)`

`f'(x) = -(1 + tan^2 x)/(tan^2 x) => f'(x) = -1 - 1/(tan^2 x)`

Since `f'(x) < 0` for any x, hence, there are no critical values of the function.

You need to check if there exists inflection points, suhc that:

`f''(x) = (-1 - 1/(tan^2 x))' => f''(x) = 1*2tan x(1 + tan^2 x)/(tan^4 x)`

Reducing duplicate factors yields:

`f''(x) = 2(1 + tan^2 x)/(tan^3 x)`

You should notice that there exists inflection points at `x in (0,kpi)` .

You need to evaluate if there exists horizontal asymptotes such that:

`lim_(x->oo)(1/tan x) = 0`

Hence, there exists horizontal asymptote at `y = 0` , thus x axis represents the horizontal asymptote.

Sketching the graph of the function yields:

The graph of `1/tan(x)` is:

The graph is periodic with periodicity `pi` . At all points where `x = n*pi + pi/2` the value of y is 0 and there is a vertical asymptote at all points where x tends to `n*pi` .