How do you evaluate the expression #cot(-pi)#?

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Answer 1 · 2016-09-14T05:07:50

See explanation.

Use #cos (-pi)==1 and sin (-pi)=0#.

Importantly, cot x is discontinuous at #x=-pi#.

Let #y = cot x = cos x/sin x, x in Q_2(-pi, -3/2pi)#, wherein it is

continuous.

As #x to - pi_+#, the approach to #-pi# is from higher values in

#Q_2#, at which #cot x < 0#, with x in #Q_2#. So, the limit, through

negative values of cot x = cos x/sin x is #-oo#

In brief, the limit #x to -pi_+# of cot x is# -oo#

By a similar argument, as #x to -pi_-# of cot x in #Q_3#, wherein

#cot x > 0#, is #oo#. .