How do you evaluate $sec [arcsin (- \frac{\sqrt{3}}{2})]$ $sec[arcsin(-sqrt3/2)]$ ?

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Answer 1 · 2015-06-13T05:18:12

Find $sec (arcsin (- \frac{\sqrt{3}}{2}))$ $sec (arcsin (-sqrt3/2))$

On the trig unit circle,

sin x = (-sqrt3/2) --> x = 240 deg

cos 240 = cos (120) = - 1/2

sec 240 = -2

How do you evaluate sec[arcsin(−√32)]sec[arcsin(-sqrt3/2)]?