How do you find #cot(-180)#?

Question

6035 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-03T09:31:20

The answer is: it does not exists.

The function #y=cotx# would be writtend in this form:

#cotx=1/tanx=1/(sinx/cosx)=cosx/sinx#.

So:

#cot(-180°)=cos(-180°)/sin(-180°)#.

#cos(-180°)=-1#,
#sin(-180°)=0#.

#cot(-180°)=1/0# that does not exists (or we can say that the function #rarr+-oo#).