How do you evaluate #cot(pi/2)#?

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Answer 1 · 2018-06-05T16:38:27

#cot(1/2pi)=0#

We know that

#cottheta=costheta/sintheta#

so

#cot(1/2pi)=cos(1/2pi)/sin(1/2pi)=0/1=0#

Answer 2 · 2018-06-05T16:41:22

#0#

Since #cot(x)=cos(x)/sin(x)#
we get
#cot(pi/2)=cos(pi/2)/sin(pi/2)#
and #cos(pi/2)=0# and #sin(pi(2)=1#
we get #0# as the searched result.