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How do you evaluate #sec((2pi)/3)-tan((pi)/6)#?

1 Answer

May 24, 2016

# -(2 + sqrt3/3)#

Explanation:

Trig table -->
#sec ((2pi)/3) = 1/cos ((2pi)/3) = 1/(-1/2) = -2#
#tan (pi/6) = sqrt3/3#
Therefor,
#sec ((2pi)/3) - tan (pi/6) = - 2 - sqrt3/3#

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