How do you evaluate #sec^2(18) - tan^2(18)#?

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May 21, 2018

#sec^2(18) - tan^2(18)#

# = 1/cos^2(18) - sin^2(18)/cos^2(18)#

# = (1-sin^2(18))/cos^2(18)#

# = cos^2(18)/cos^2(18)#

# = 1#

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