How do you express $sec(-1290^circ)$ as a trig function of an angle in Quadrant I?

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Answer 1 · 2018-08-09T02:36:49

$- sec 30^o = - 2/ sqrt3 = -1.1547$ , nearly.

$sec ( - 1290^o ) = sec ( - 7 ( 180^o ) - 30^o )$

$rArr$ - 1290^o in Q_2#, and so,

$sec ( - 1290^o ) = - sec ( - 30^o ) = - sec 30^o = - 2/ sqrt3$

$= -1.1547$ , nearly.

How do you express sec(-1290^circ)sec(-1290^circ) as a trig function of an angle in Quadrant I?