Describe the graph of `y=1/(tanx)` :

This is the same as `y=cotx` .(`1/(tanx)=1/(sinx/cosx)=cosx/sinx=cotx` )

The domain is `x!= 0+-npi,n in ZZ` (n an integer). The vertical asymptotes occur at `x=0+-npi` .

The range is all reals.

There is no y-intercept.

The zeros are at `x=pi/2 +- npi,n in ZZ`

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Describe the graph of `y=1/(tanx)` :

This is the same as `y=cotx` .(`1/(tanx)=1/(sinx/cosx)=cosx/sinx=cotx` )

The domain is `x!= 0+-npi,n in ZZ` (n an integer). The vertical asymptotes occur at `x=0+-npi` .

The range is all reals.

There is no y-intercept.

The zeros are at `x=pi/2 +- npi,n in ZZ`

The graph:

You need to remember that the domain of tangent function is `(-pi/2,pi/2)` and since tangent function is denominator of the given function, its value needs to be different from zero, hence, the domain of the given function is `(-pi/2,pi/2) - {0}` and the range of the given function is R.

You need to find the vertical asymptote of the function such that:

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

**Hence, evaluating the vertical asymptote yields `x = 0.` **

You need to find the zeroes of the function such that:

`1/tan x = 0` invalid for any value of `x in (-pi/2,pi/2)-{0}`

**Since the value `x = 0` is excluded from domain of function, hence, the function has no y intercepts.**