How do you evaluate #sec((11pi)/2)#?

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How do you evaluate #sec((11pi)/2)#?

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Answer 1 · 2017-02-16T01:31:43

undefined infinity

Explanation:

#sec ((11pi)/2) = 1/cos ((11p)/2)#
#cos ((11pi)/2) = cos ((3pi)/2 + (8pi)/2) = cos ((3pi)/2 + 4pi) = #
#= cos ((3pi)/2) = 0#
Finally:
#sec ((11pi)/2) = 1/(cos) = 1/0 #--> undefined infinity

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How do you evaluate #sec((11pi)/2)#?

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