Question

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

Answer 1

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

Explanation:

`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.