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

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Answer 1 · 2016-05-24T04:41:41

$-(2 + sqrt3/3)$

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

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