How do you simplify the expression $sec(arccos (-3/5))$ ?

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Answer 1 · 2016-07-04T14:13:16

$sec(arccos(-3/5))=-5/3$

Let $arccos(-3/5)=theta$

Therefore $costheta=-3/5$

and hence $sec(arccos(-3/5))=sectheta$

= $1/costheta=1/(-3/5)$

= $1xx-5/3=-5/3$

Answer 2 · 2016-07-04T16:28:05

$-5/3$

Use cos arc cos x = x.

Here, $sec arc cos (-3/5)=1/(cos arc cos (-3/5))=1/(-3/5)=-5/3$

How do you simplify the expression sec(arccos (-3/5))sec(arccos (-3/5))?