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How do you find the value of #cot (pi/2)#?

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Jun 9, 2015

You can use two trig identities: #cotx = 1/tanx# and #tanx = (sinx)/(cosx)#.

#cotx = 1/tanx = cosx/sinx#

#pi/2 "rad" = 90^o#

#sin 90^o = 1#
#cos 90^o = -1#

#cot90^o = (cos90^o)/(sin90^o) = (-1)/(1) = -1#

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