What is the exact value of #cot(-90)# and #csc690#, using unit circle?

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Answer 1 · 2017-10-23T00:18:47

It depends on what you do want to use.

I assume that you actually mean to ask about #cot (-90^@)#

(There is no nice way to express the exact value of #cot (-90)#).

If you know the following, then there is no need to use the unit circle:

#cot A = cosA/sinA#

#sin(-A) = -sin(A)# and #cos(-A) = cos(A)#

#sin(90^@) = 1# and #cos(90^@) = 0#

We get
#cot (90^@) = cos(-90^@)/sin(-90^@) = cos(90^@)/(-sin(90^@)) = (0)/(-(1))=0#

For the other we can use

#csc(690^@) = csc(690^@ - 360^@) = csc(330^@)#

# = csc(330^@-360^@) = csc(-30^@)#

# = 1/sin(-30^@) = 1/(-sin(30^@)) = 1/(-(1/2)) = -2#